\chapter{Stability of Nonlinear Systems}
\label{ch:stability}
Our goal in this chapter is to extend concepts introduced in Chapter~\ref{chapter:instab}, to nonlinear time-varying dynamic system models described by
\[
\dot{\mb{\bm$x$}}(t)=\mb{\bm$f$}(t,\mb{\bm$x$}(t)), \qquad \mb{\bm$x$}(t_0)=\mb{\bm$x$}_0,
\]
where recall $\mb{\bm$x$}(t) \in \mb{\Bb R}^n$ is the state vector, and \mb{\bm$f$} is a vector-valued function.\\
To proceed with our presentation we need the following lemma.
\begin{lemma}
\label{lemma:con_kap}
For a continuously differentiable function $V : \mb{\Bb R}^n \to \mb{\Bb R}$, where $V(\mb{\bm$0$}) = 0$, we have $V( \mb{\bm$x$} ) > 0$ for all $\mb{\bm$x$} \in {\cal S} \setminus \{ \mb{\bm$0$} \}$, if and only if there is a continuous, strictly increasing, function $\psi : \mb{\Bb R} \to \mb{\Bb R}$ such that $\psi (0) = 0$, $\psi (r) > 0$ for $r>0$, and $V(\mb{\bm$x$}) \ge \psi (r)$, where $r=\| \mb{\bm$x$} \|$, and $r \le R = \max \{ q : B(\mb{\bm$0$}, q)  \subseteq {\cal S}\}$. Here $B(\mb{\bm$0$}, q )$ denotes an open ball with the center at $\mb{\bm$0$}$ and radius $q$.
\end{lemma}
{\bf Proof}\quad Sufficiency $(\,\Leftarrow\,)$ is clear.\\
To prove necessity $(\, \Rightarrow \,)$ define the following function $\kappa : \mb{\Bb R} \to \mb{\Bb R}$ as
\[
\kappa (y) = \min_{\stackrel{\mb{\bm$x$} \in {\cal S}}{\| \mb{\bm$x$}\| = y}} V(\mb{\bm$x$} )
\]
Note that the function $\kappa$ is well defined since the set
\[
\{ \mb{\bm$x$} : \|\mb{\bm$x$} \| = y \} \bigcup {\cal S}
\]
is closed and bounded, and hence compact.

We shall now show that $\kappa$ is continuous on $B(\mb{\bm$0$}, R )$. Then we will construct an appropriate function $\psi$ satisfying the hypotheses of the lemma. To show that $\kappa$ is continuous on $B(\mb{\bm$0$}, R )$ we shall use the fact that $V$ is uniformly continuous in ${\cal S}$. To this end let $\mb{\bm$x$}^{(1)} \in {\cal S}$ be such that $\|\mb{\bm$x$}^{(1)}\| = y$ and $\kappa (y) = V \left(\mb{\bm$x$}^{(1)} \right)$. Next, let $\Delta y$ be an increment of $y$ and $\mb{\bm$x$}^{(2)} \in {\cal S}$ be a vector such that $y+\Delta y = \|\mb{\bm$x$}^{(2)}\|$. Observe that
\[
V \left(\mb{\bm$x$}^{(2)} \right) \ge \kappa ( y + \Delta y).
\]
Choose now a vector $\mb{\bm$x$}^{(3)}$ such that
\[
\|\mb{\bm$x$}^{(3)}\| = y + \Delta y,\qquad \mb{and} \qquad V\left(\mb{\bm$x$}^{(3)} \right)= \kappa (y + \Delta y).
\]
Note that for any vector, say $\mb{\bm$x$}^{(4)}$, such that $\|\mb{\bm$x$}^{(4)}\|=y$ we have $V\left( \mb{\bm$x$}^{(4)}\right) \ge \kappa(y)$---see Figure~\ref{fig:con_kap}.
\begin{figure}
\vspace{3in}
\hskip1in\special{bmp:c:/mybook/con_kap.bmp x=4in y=3in}
\caption{A two-dimensional illustration of the proof of continuity of the function $\kappa$.}
\label{fig:con_kap}
\end{figure}
By the uniform continuity of $V$, for any $\epsilon > 0$ there exists a $\delta > 0$ such that if
\[
\|\mb{\bm$x$}^{(1)}-\mb{\bm$x$}^{(2)}\| < \delta \quad \mb{and} \quad \|\mb{\bm$x$}^{(3)}-\mb{\bm$x$}^{(4)}\| < \delta,
\]
then
\[
\left\arrowvert V\left(\mb{\bm$x$}^{(1)} \right)-V\left(\mb{\bm$x$}^{(2)} \right)\right\arrowvert < \epsilon \quad \mb{and} \quad \left\arrowvert V\left(\mb{\bm$x$}^{(3)} \right)-V\left(\mb{\bm$x$}^{(4)} \right)\right\arrowvert < \epsilon.
\]
From $\left\arrowvert V\left(\mb{\bm$x$}^{(1)} \right)-V\left(\mb{\bm$x$}^{(2)} \right)\right\arrowvert < \epsilon$ we obtain
\[
-\epsilon < V\left(\mb{\bm$x$}^{(1)} \right)-V\left(\mb{\bm$x$}^{(2)} \right) < \epsilon.
\]
Thus
\[
V\left(\mb{\bm$x$}^{(1)} \right) > V\left(\mb{\bm$x$}^{(2)} \right) - \epsilon.
\]
From $\left\arrowvert V\left(\mb{\bm$x$}^{(3)} \right)-V\left(\mb{\bm$x$}^{(4)} \right)\right\arrowvert < \epsilon$ we get
\[
V\left(\mb{\bm$x$}^{(3)} \right)>V\left(\mb{\bm$x$}^{(4)} \right)- \epsilon.
\]
From the definition of the function $\kappa$ it follows that
\[
V \left( \mb{\bm$x$}^{(2)} \right) \ge V \left( \mb{\bm$x$}^{(3)} \right)
\]
and
\[
V \left( \mb{\bm$x$}^{(4)} \right) \ge V \left( \mb{\bm$x$}^{(1)} \right).
\]
Since $V\left(\mb{\bm$x$}^{(1)} \right) > V\left(\mb{\bm$x$}^{(2)} \right) - \epsilon$, we obtain
\[
V\left(\mb{\bm$x$}^{(1)} \right) > V\left(\mb{\bm$x$}^{(3)} \right) - \epsilon
\]
and therefore
\[
- \epsilon < V\left(\mb{\bm$x$}^{(1)} \right) - V\left(\mb{\bm$x$}^{(3)} \right).
\]
Since $V\left(\mb{\bm$x$}^{(3)} \right) > V\left(\mb{\bm$x$}^{(4)} \right) - \epsilon$,
\[
V\left(\mb{\bm$x$}^{(3)} \right) > V\left(\mb{\bm$x$}^{(1)} \right) - \epsilon,
\]
and thus
\[
V\left(\mb{\bm$x$}^{(1)} \right) - V\left(\mb{\bm$x$}^{(3)} \right) < \epsilon.
\]
The above can be represented as
\[
\left\arrowvert V\left(\mb{\bm$x$}^{(1)} \right) - V\left(\mb{\bm$x$}^{(3)} \right)\right\arrowvert < \epsilon,
\]
or equivalently as
\[
|\kappa(y)- \kappa(y + \Delta y)| < \epsilon
\]
if
\[
\Delta y < \delta
\]
which proves the continuity of $\kappa$.


We are now ready to construct a function $\psi$ that satisfies the hypotheses of the lemma. It has the form
\[
\psi (r) = r \min _{r \le y \le R} \frac{\kappa(y)}{y}.
\]
Note that $\psi$ is continuous, $\psi (0) = 0$, and it is positive and strictly increasing---see Figure~\ref{fig:fun_psi} for a graphical illustration of the way $\psi$ is constructed from $\kappa$.
\vspace*{0.1in}
\begin{figure}
\vspace{3in}
\hskip1in\special{bmp:c:/mybook/fun_psi.bmp x=4in y=3in}
\caption{Constructing the function $\psi$.}
\label{fig:fun_psi}
\end{figure}
The proof of the lemma is thus complete.\\
\hspace*{5.5in}$\Box$


We can characterize a positive definite function $V$ in another way as in the following lemma.
\begin{lemma}
A function $V : \mb{\Bb R}^n \to \mb{\Bb R}$ is positive definite if and only if there exists a continuous nondecreasing function $\alpha : \mb{\Bb R} \to \mb{\Bb R}$, such that $\alpha (0) = 0$, $\alpha (r) > 0$ for $r>0$, and $V(\mb{\bm$x$}) \ge \alpha(r)$, where $r=\|\mb{\bm$x$}\|$, and $r \le R = \max \{q : B(\mb{\bm$0$},q) \subseteq {\cal S}\}$.
\end{lemma}
A possible choice for $\alpha$ is
\[
\alpha(r)=\min_{r \le \|\mb{\bm$x$}\| \le R} V(\mb{\bm$x$}).
\]
Note that the function $\alpha$ is nondecreasing because as $r$ increases, the minimum is taken over a smaller region. Furthermore, $\alpha(r) > 0$ whenever $r>0$ since $V(\mb{\bm$x$}) > 0$ for $\mb{\bm$x$} \in {\cal S} \setminus \{\mb{\bm$0$}\}$. The function $\alpha$ is continuous since $V$ is continuous.

So far we only considered positive definite, or negative definite functions, that are time-invariant. We now extend our definition of positive definiteness to time-varying functions of the form $W(t,\mb{\bm$x$})$. We first define a time-varying positive semidefinite function.
\begin{defn}
A real-valued function $W : \mb{\Bb R} \times \mb{\Bb R}^n \to \mb{\Bb R}$ is positive semidefinite in a region ${\cal S} \subseteq \mb{\Bb R}^n$ if for all $t$ and all $\mb{\bm$x$}$ in $\cal S$
\begin{itemize}
\item[(i)] $W$ is continuous and has continuous partial derivatives with respect to all of its arguments, that is $W \in {\cal C}^1$,
\item[(ii)] $W(t,\mb{\bm$0$})=0$,
\item[(iii)] $W(t,\mb{\bm$x$}) \ge 0$.
\end{itemize}
\end{defn}
\begin{defn}
A real-valued function $W : \mb{\Bb R} \times \mb{\Bb R}^n \to \mb{\Bb R}$ is positive definite in a region ${\cal S} \subseteq \mb{\Bb R}^n$ if for all $t$ and all $\mb{\bm$x$}$ in $\cal S$
\begin{itemize}
\item[(i)] $W \in {\cal C}^1$,
\item[(ii)] $W(t,\mb{\bm$0$})=0$,
\item[(iii)] there exists a positive definite time-invariant function $V$ such that $W(t,\mb{\bm$x$}) \ge V(\mb{\bm$x$})$.
\end{itemize}
\end{defn}
We next introduce a notion of a {\em decrescent\/} function.
\begin{defn}
A time-varying function $W : \mb{\Bb R} \times \mb{\Bb R}^n \to \mb{\Bb R}$ is decrescent in a region $\cal S$ if a positive definite function $\tilde{V}$ exists such that for all $\mb{\bm$x$}$ in $\cal S$ and all $t$
\[
|W(t,\mb{\bm$x$})| \le \tilde{V}(\mb{\bm$x$}),
\]
that is, $W$ tends to zero uniformly with respect to $t$ as $\| \mb{\bm$x$} \| \to 0$.
\end{defn}

Consider now a dynamic system modeled by $\dot{\mb{\bm$x$}}(t)=\mb{\bm$f$}(t,\mb{\bm$x$}(t))$ subject to the initial condition $\mb{\bm$x$}(t_0)=\mb{\bm$x$}_0$. Suppose that the system has an isolated equilibrium point at $\mb{\bm$x$}_{eq}$, that is, $\mb{\bm$f$}(t,\mb{\bm$x$}_{eq})=\mb{\bm$0$}$ for all $t$. Let $W(t,\mb{\bm$x$})$ be a time-dependent real-valued function whose time derivative along the solution $\mb{\bm$x$}(t)=\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)$ of the given differential equation, is
\[
\frac{dW(t,\mb{\bm$x$})}{dt}=\dot{W}(t,\mb{\bm$x$})=\frac{\partial W}{\partial t}+\left(\nabla W \right)^T \dot{\mb{\bm$x$}}.
\]
Then, we have the following theorem concerning the stability of the equilibrium point in the sense of Lyapunov.
\begin{thm}
\label{thm:loc_stab}
If in some neighborhood $\cal S$ of the equilibrium state $\mb{\bm$x$}_{eq}$, there is a positive definite function $W$ such that its time derivative along any solution of $\dot{\mb{\bm$x$}}(t)=\mb{\bm$f$}(t,\mb{\bm$x$}(t))$ is negative semidefinite in $\cal S$, then the equilibrium state $\mb{\bm$x$}_{eq}$ is stable. If $W$ is also decrescent, then the equilibrium state is uniformly stable.
\end{thm}
{\bf Proof}\quad Without loss of generality we assume that $\mb{\bm$x$}_{eq}=\mb{\bm$0$}$. Since $W$ is positive definite, by Lemma~\ref{lemma:con_kap} there is a strictly increasing function $\psi _1 (r)$, $\psi (0)=0$ such that
\[
W(t,\mb{\bm$x$}) \ge \psi _1 (\|\mb{\bm$x$}\|),
\]
where $\|\mb{\bm$x$}\| \le R$, and $R$ is the radius of the largest ball contained in $\cal S$.\\
Let
\[
\psi _2 (t_0,r)= \max _{\|\mb{\bm$x$}\| \le r} W(t_0,\mb{\bm$x$}).
\]
Then
\[
\psi _1 (r) \le \psi _2 (t_0,r).
\]
Thus, for any $\epsilon > 0$ there is a positive $\delta$ that depends on $\epsilon$ and $t_0$, such that
\[
\psi _2 (t_0, \delta ) < \min \{ \psi _1 ( \epsilon ), \psi _1 (R) \}.
\]
Figure~\ref{fig:fin_del} illustrates how to determine a $\delta$ given an $\epsilon$.
\begin{figure}
\vspace{3in}
\hskip1in\special{bmp:c:/mybook/fin_del.bmp x=4in y=3in}
\caption{Determining a $\delta$ given an $\epsilon$.}
\label{fig:fin_del}
\end{figure}
By assumption $\frac{dW(t,\mb{\bm$x$})}{dt} \le 0$ along any solution of the system. Hence $W(t,\mb{\bm$x$})$ is a non-increasing function and therefore
\[
W(t,\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)) \le W(t_0,\mb{\bm$x$}_0) \qquad \mb{for all} \qquad t \ge t_0.
\]
If $\|\mb{\bm$x$}_0\| < \delta$, then for $t \ge t_0$ we have
\[
\psi _1(\epsilon)>\psi_2(t_0, \delta) \ge W(t_0,\mb{\bm$x$}_0) \ge W(t,\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)) \ge \psi_1(\|\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)\|).
\]
Thus
\[
\psi_1 (\epsilon) > \psi_1(\|\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)\|).
\]
Since $\psi_1$ is strictly increasing, we have to have
\[
\|\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)\| < \epsilon
\]
for all $t \ge t_0$. This means that the null solution is (locally) stable which completes the proof of the first part of the theorem.


If $W$ is also decrescent, then there is a nondecreasing function $\tilde{\psi}_2:\mb{\Bb R} \to \mb{\Bb R}$, $\tilde{\psi}_2(0)=0$, and
\[
W(t,\mb{\bm$x$}) \le \tilde{\psi}_2 (\|\mb{\bm$x$}\|)
\]
for all $\mb{\bm$x$}$ such that $\|\mb{\bm$x$}\| \le R$ and all $t$. Let
\[
\tilde{\psi}_2 (\delta ) < \min \{\psi_1 (\epsilon ), \psi_1 ( R )\}.
\]
Note that $\delta$ depends now only on $\epsilon$, and not on $t_0$ as before. To show the uniform stability of the null solution, we repeat the argument above with $\tilde{\psi}_2$ replacing $\psi_2$. This completes the proof of the theorem.\\
\hspace*{5.5in} $\Box$


The next theorem gives us a sufficiency condition for an isolated equilibrium point to be uniformly asymptotically stable.
\begin{thm}
\label{thm:uniform_as}
If in some neighborhood $S$ of the given isolated equilibrium point $\mb{\bm$x$}_{eq}$ there is a positive definite and decrescent function $W$ such that its time derivative along any solution of the system is negative definite in $S$, then the equilibrium state is uniformly asymptotically stable. 
\end{thm}
{\bf Proof}\quad As before, without loss of generality we assume that $\mb{\bm$x$}_{eq}= \mb{\bm$0$}$. The previous theorem insures uniform stability of the equilibrium state $\mb{\bm$x$}_{eq}$. Thus, only the uniform convergence remains to be shown. From the assumptions of the theorem it follows that there exist strictly increasing functions $\psi_1$ and $\psi_3$, and a non-decreasing function $\psi_2$ such that
\begin{eqnarray*}
\psi_1 (0) &=& \psi_2 (0) = \psi_3 (0),\\
\psi_1(\|\mb{\bm$x$}\|) &\le& W(t,\mb{\bm$x$}) \le \psi_2(\|\mb{\bm$x$}\|),\\
\psi_3(\|\mb{\bm$x$}\|) &\le& - \dot{W}(t,\mb{\bm$x$}),
\end{eqnarray*}
for all $t$ and all $\mb{\bm$x$}$ such that $\|\mb{\bm$x$}\| \le R$. Given an $\epsilon_1$. We define positive constants $\delta_1$, $\delta_2$, and $T$ as follows:
\begin{eqnarray*}
\psi_2(\delta_1) &<& \psi_1(R),\\
\psi_2(\delta_2) &<& \min \{\psi_1 (\epsilon_1 ), \psi_2 (\delta _1) \},\\
T&=&\psi_1 (R)/\psi_3(\delta _2)
\end{eqnarray*}
Note that $\delta_1$ is independent of $t_0$, and $T$ depends on $\epsilon_1$ and $\delta_1$.\\
We shall prove the theorem in two stages. First, we shall show that for some $t_2 \in \left[ t_0,t_0+T\right]$,
\[
\|\mb{\bm$x$}(t_2;t_0,\mb{\bm$x$}_0)\| < \delta_2
\]
whenever $\|\mb{\bm$x$}_0 \|< \delta_1$. Then, we shall show that once $\|\mb{\bm$x$}(t_2;t_0,\mb{\bm$x$}_0) \| < \delta_2$, we have 
\[
\|\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0) \| < \epsilon_1 \quad \mb{for all} \quad t \ge t_2.
\]
We begin with proving the claim that for some $t_2 \in \left[ t_0,t_0+T\right]$,
\[
\|\mb{\bm$x$}(t_2;t_0,\mb{\bm$x$}_0)\| < \delta_2
\]
whenever $\|\mb{\bm$x$}_0 \|< \delta_1$. We prove this claim by contradiction. We thus assume that for all $t\in \left[t_0,t_0+T\right]$,
\[
\|\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)\| \ge \delta_2
\]
if $\|\mb{\bm$x$}_0 \|< \delta_1$. We thus have
\begin{eqnarray*}
0 < \psi_1 (\delta _2 ) &\le& W[t_0+T,\mb{\bm$x$}(t_0+T;t_0,\mb{\bm$x$}_0)]\\
&=& W(t_0,\mb{\bm$x$}_0) + \int_{t_0}^{t_0+T} \dot{W}[t,\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)]dt\\
&\le& W(t_0,\mb{\bm$x$}_0) - \int_{t_0}^{t_0+T} \psi_3(\|\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)\|)dt\\
&\le& W(t_0,\mb{\bm$x$}_0) - T\psi_3(\delta_2)\\
&\le& \psi_2(\|\mb{\bm$x$}_0\|)-T\psi_3(\delta_2)\\
&<& \psi_2(\delta_1)-\psi_1(R)\\
&<&0,
\end{eqnarray*}
which is a contradiction, and thus we proved the first part of the theorem. We now proceed with the second part. From the first part we have that if $\|\mb{\bm$x$}_0\| < \delta_1$, then for some $t_2 \in \left[t_0,t_0+T\right]$ we have $\|\mb{\bm$x$}(t_2;t_0,\mb{\bm$x$}_0)\| < \delta_2$. For $t \ge t_2$ we have
\[
\psi_1(\|\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)\|) \le W\left(t,\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)\right) \le W\left(t_2,\mb{\bm$x$}(t_2;t_0,\mb{\bm$x$}_0)\right) < \psi_2(\delta_2)<\psi_1(\epsilon_1).
\]
Hence
\[
\psi_1(\|\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)\|)<\psi_1(\epsilon_1),
\]
and since $\psi_1$ is strictly increasing,
\[
\|\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)\|< \epsilon_1
\]
for all $t \ge t_2$ which proves the uniform convergence. The proof of the theorem is thus complete.\\
\hspace*{5.5in} $\Box$


The above two theorems are concerned with local stability. We now proceed to prove sufficiency conditions for global stability. For this we need the following definitions. We first define a {\em radially unbounded\/} or {\em coercive\/} function.
\begin{defn}
A time-varying function $W : \mb{\Bb R} \times \mb{\Bb R}^n \to \mb{\Bb R}$ is radially unbounded, or coercive, if as $\|\mb{\bm$x$}\| \to \infty$, the function $W$ tends to infinity uniformly with respect to $t$, that is, for any given positive real number $M$, there is a positive real number $N$ so that $W(t,\mb{\bm$x$}) > M$ for all $t$ and all $\mb{\bm$x$}$ such that $\|\mb{\bm$x$}\| > N$.
\end{defn}
The next two definitions are concerned with {\em Lagrange stable equilibrium states}.
\begin{defn}
\label{defn:lagr_stab}
The equilibrium state $\mb{\bm$x$}_{eq}$ is called Lagrange stable, or bounded, if for every real number $d>0$ there exists a positive real number $b=b(t_0,d)$ such that if
\[
\| \mb{\bm$x$}_0 - \mb{\bm$x$}_{eq} \| < d
\]
then
\[
\| \mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)-\mb{\bm$x$}_{eq} \| < b
\]
for all $t \ge t_0$.
\end{defn}
\begin{defn}
The equilibrium state $\mb{\bm$x$}_{eq}$ is uniformly bounded if the number $b$ in Definition~\ref{defn:lagr_stab} can be taken independent of the initial time $t_0$.
\end{defn}
We now define {\em uniformly asymptotically stable in the large\/} equilibrium state.
\begin{defn}
If the equilibrium state $\mb{\bm$x$}_{eq}$ is uniformly stable, uniformly bounded, and all trajectories are uniformly convergent to $\mb{\bm$x$}_{eq}$, then this equilibrium state is called uniformly asymptotically stable in the large, or globally uniformly asymptotically stable.
\end{defn}
The following theorem gives a sufficiency condition for an equilibrium state to be globally uniformly asymptotically stable.
\begin{thm}
\label{thm:guas}
If there is a function $W$ that is positive definite for all $\mb{\bm$x$} \ne \mb{\bm$x$}_{eq}$, $W(t,\mb{\bm$x$}_{eq})=0$, decrescent, radially unbounded, and if its time derivative along any solution of the system is negative definite for all $\mb{\bm$x$}$, then the equilibrium state $\mb{\bm$x$}_{eq}$ is globally uniformly asymptotically stable. 
\end{thm}
{\bf Proof}\quad By Theorem~\ref{thm:loc_stab}, the equilibrium state $\mb{\bm$x$}_{eq}$ is uniformly stable. Similarly as in Theorem~\ref{thm:uniform_as} we construct the functions $\psi_1$, $\psi_2$, and $\psi_3$. Since, by the assumption, $W$ is radially unbounded, $\psi_1(\|\mb{\bm$x$}\|) \to \infty$ as $\|\mb{\bm$x$}\| \to \infty$. Therefore, for any constant $d>0$, there is a constant $b > 0$ such that $\psi_1(b) > \psi_2(d)$. See Figure~\ref{fig:fin_bpr} for an illustration of the way a constant $b$ is found.
\begin{figure}
\vspace{3in}
\hskip1in\special{bmp:c:/mybook/fin_bpr.bmp x=4in y=3in}
\caption{Determining a constant $b>0$ given $d>0$.}
\label{fig:fin_bpr}
\end{figure}
If $\| \mb{\bm$x$}_0 \|<d$, then $\| \mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0) \|<b$ for all $t \ge t_0$. Indeed, since
\[
\psi_1(b)> \psi_2(d) \ge W(t_0,\mb{\bm$x$}_0) \ge W(t,\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)) \ge \psi_1(\| \mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0) \|),
\]
\[
\psi_1(b)>\psi_1(\| \mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0) \|).
\]
The above implies
\[
\| \mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0) \|<b
\]
since $\psi_1$ is strictly increasing. Hence all trajectories of the system are uniformly bounded.\\
To prove their convergence to the equilibrium state, we define $\delta$ and $T$, given $\epsilon_1$, as follows
\begin{eqnarray*}
\psi_2(\delta_2) &<& \psi_1(\epsilon_1),\\
T &=& \psi_1(b)/ \psi_3(\delta).
\end{eqnarray*}
If $\|\mb{\bm$x$}_0\|<d$, then for all $t>t_0+T$ we have
\[
\|\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)\|<\epsilon_1,
\]
which can be shown using the same arguments as in the proof of Theorem~\ref{thm:uniform_as} with $b$ playing the role of $R$, and $d$ playing the role of $\delta_1$. This proves that all trajectories of the system converge uniformly to the equilibrium state. Hence the equilibrium state is globally uniformly asymptotically stable.\\
\hspace*{5.5in} $\Box$


We now adapt the Lyapunov method to test for instability of nonlinear dynamic systems.
\begin{thm}
Let $\mb{\bm$x$}_{eq}$ be an isolated equilibrium state of the system $\dot{\mb{\bm$x$}}=\mb{\bm$f$}(t,\mb{\bm$x$})$, that is $\mb{\bm$f$}(t,\mb{\bm$x$}_{eq})=\mb{\bm$0$}$. If there exists a function $W:\mb{\Bb R} \times \mb{\Bb R}^n \to \mb{\Bb R}$ such that in the region ${\cal S}=\{ \mb{\bm$x$}:\|\mb{\bm$x$}-\mb{\bm$x$}_{eq}\| \le R \}$
\begin{itemize}
\item[(i)]
$W \in {\cal C}^1$,
\item[(ii)]
$W(t,\mb{\bm$x$}_{eq})=0$,
\item[(iii)]
$\dot{W}(t,\mb{\bm$x$}) > 0$ for all $\mb{\bm$x$} \in {\cal S} \setminus \{\mb{\bm$x$}_{eq} \}$,
\item[(iv)]
in every neighborhood of $\mb{\bm$x$}_{eq}$ in ${\cal S}$, for any $t_0$ there is at least one point $\mb{\bm$x$}$ at which $W(t_0,\mb{\bm$x$})>0$,
\end{itemize}
then the equilibrium state $\mb{\bm$x$}_{eq}$ is unstable.
\end{thm}
{\bf Proof}\quad Without loss of generality we assume that $\mb{\bm$x$}_{eq}=\mb{\bm$0$}$. By assumption~{\it (iv)} for any $t_0$ and any $\delta$, $0<\delta<R$ there exists $\mb{\bm$x$}_{\delta}$, $0<\|\mb{\bm$x$}_{\delta}\| < \delta$ such that $W(t_0,\mb{\bm$x$}_{\delta}) > 0$. The system trajectory $\mb{\bm$x$}(t)=\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_{\delta})$ cannot approach the equilibrium state $\mb{\bm$x$}_{eq}=\mb{\bm$0$}$ as $t \to \infty$ since $W(t,\mb{\bm$0$})=0$ but $W(t_0,\mb{\bm$x$}_{\delta}) > 0$ and $\dot{W}(t,\mb{\bm$x$})>0$ for $\mb{\bm$x$} \ne \mb{\bm$x$}_{eq}$. Since $\dot{W} \in {\cal C}^1$ and the trajectory is bounded away from the equilibrium state---see Figure~\ref{fig:instabt}.
\begin{figure}
\vspace{3in}
\hskip1in\special{bmp:c:/mybook/instabt.bmp x=4in y=3in}
\caption{Illustration of the Instability Test.}
\label{fig:instabt}
\end{figure}
This means that there is a constant $m>0$ such that for all $t \ge t_0$
\[
\dot{W}(t,\mb{\bm$x$}(t)) \ge m >0
\]
as long as
\[
\|\mb{\bm$x$}(t)\| \le R, \qquad \mb{\bm$x$} \ne \mb{\bm$0$}.
\]
Hence
\[
W(t,\mb{\bm$x$})=W(t_0,\mb{\bm$x$}_{\delta})+\int_{t_0}^{t} \dot{W}(s,\mb{\bm$x$}(s))ds \ge W(t_0,\mb{\bm$x$}_{\delta})+m(t-t_0).
\]
The function $W$ is continuous and therefore bounded on the compact set ${\cal S}$. However, the right hand side of the above is unbounded. Therefore the trajectory $\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_{\delta})$ cannot remain infinitely in $\cal S$. Thus for some $t_1>t_0$ the trajectory reaches the boundary of the region $\|\mb{\bm$x$}\| \le R$. Hence given any $\epsilon$, $0<\epsilon<R$, then for any $\delta>0$ no matter how small there is at least one trajectory such that $\|\mb{\bm$x$}_{\delta}\| < \delta$ but $\|\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_{\delta})\| > \epsilon$ for some $t_1 > t_0$. Therefore the equilibrium state is unstable.\\
\hspace*{5.5in} $\Box$


\begin{corollary}
If there exists a function $W:\mb{\Bb R} \times \mb{\Bb R}^n \to \mb{\Bb R}$ such that in the region ${\cal S}=\{ \mb{\bm$x$}:\|\mb{\bm$x$}-\mb{\bm$x$}_{eq}\| \le R \}$
\begin{itemize}
\item[(i)]
$W \in {\cal C}^1$,
\item[(ii)]
$W(t,\mb{\bm$x$}_{eq})=0$,
\item[(iii)]
$\dot{W}(t,\mb{\bm$x$}) < 0$ for all $\mb{\bm$x$} \in {\cal S} \setminus \{\mb{\bm$x$}_{eq} \}$,
\item[(iv)]
in every neighborhood of $\mb{\bm$x$}_{eq}$ in ${\cal S}$, for any $t_0$ there is at least one point $\mb{\bm$x$}$ at which $W(t_0,\mb{\bm$x$})<0$,
\end{itemize}
then the equilibrium state $\mb{\bm$x$}_{eq}$ is unstable.
\end{corollary}

\section{Practical Stability}
A drawback to the discontinuous controllers discussed in the previous chapters is that rapid switching tends to excite high frequency modes of the plant and this may degrade the performance of the controller. These problems can be alleviated with the insertion of a {\em boundary layer\/} about a switching surface resulting in a continuous control strategy. However, there is a price tag for making discontinuous controllers continuous. The closed-loop system driven by a boundary layer controller may only be {\em uniformly ultimately bounded\/} rather than asymptotically stable. Specifically, we can only prove that a neighborhood of  the equilibrium point of interest, rather than the equilibrium point itself, is stable. We refer to this kind of stability as practical stability. The notions we are concerned with in this chapter are spelled out in the following three definitions.
\begin{defn}
The ball $B_{q}=\left\{ \mb{\bm$x$}\,:\, \|\mb{\bm$x$}\| \le q \right\}$ is uniformly stable if for all $\varepsilon > q$, there is $\delta=\delta(\varepsilon) > 0$ so that if $\|\mb{\bm$x$}(t_0) \| \le \delta$, then $\|\mb{\bm$x$}(t)\| \le \varepsilon$ for $t \ge t_0$.
\end{defn}
For the benefit of the reader we now rephrase the definition of uniform boundedness from Section~\ref{ch:stability}.
\begin{defn}
A system is uniformly bounded if for any given $d>0$, there exists $b=b(d)< \infty$ so that for each solution $\mb{\bm$x$}(t) = \mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)$ with $\|\mb{\bm$x$}_0\| \le d$, we have $\|\mb{\bm$x$}(t)\| \le b$ for all $t\ge t_0$.
\end{defn}
We next define the uniform ultimate boundedness of a system, or for short, practical stability.
\begin{defn}
A system is uniformly ultimately bounded, or practically stable, with respect to the ball $B$ if for every $d>0$, there is $T(d) \ge t_0$ such that for each solution $\mb{\bm$x$}(t)$ with $\|\mb{\bm$x$}(t_0)\| \le d$, we have $\mb{\bm$x$} \in B$ for $t \ge T(d)$.
\end{defn}

We now consider a dynamic system model of the form
\[
\dot{\mb{\bm$x$}}(t)=\mb{\bm$f$}(t,\mb{\bm$x$}(t))+\mb{\bm$G$}(t,\mb{\bm$x$}(t))(\mb{\bm$u$}(t)+\mb{\bm$\xi$}(t,\mb{\bm$x$}(t)),\qquad\mb{\bm$x$}(t_0)=\mb{\bm$x$}_0,
\]
where $\mb{\bm$x$}(t)\in \mb{\Bb R}^n$, $\mb{\bm$u$}(t)\in \mb{\Bb R}^m$, and $\mb{\bm$\xi$}(t,\mb{\bm$x$}(t))\in \mb{\Bb R}^m$ represents the lumped uncertainties. We assume that the uncertain element $\mb{\bm$\xi$}$ is bounded by a known continuous non-negative function $\rho$, that is,
\[
\|\mb{\bm$\xi$}(t,\mb{\bm$x$}(t))\| \le \rho(t,\mb{\bm$x$}(t)).
\]
We also assume that the equilibrium state $\mb{\bm$x$}=\mb{\bm$0$}$ of the uncontrolled nominal system $\dot{\mb{\bm$x$}}(t)=\mb{\bm$f$}(t,\mb{\bm$x$}(t))$ is globally uniformly asymptotically stable, and that we were able to construct a Lyapunov function $W:\mb{\Bb R} \times \mb{\Bb R}^n \to \mb{\Bb R}$ corresponding to this equilibrium state. The Lyapunov function $W$ we found satisfies the condition listed in Section~\ref{ch:stability}. Specifically, there exist three continuous functions $\psi_i$, $i=1,2,3$, where $\psi_1$ and $\psi_3$ are strictly increasing, and $\psi_2$ nondecreasing, such that
\begin{eqnarray*}
\psi_i(0) &=& 0,\quad i=1,2,3,\\
\lim_{r \to \infty} \psi_1(r)&=& \infty,\\
\psi_1(\|\mb{\bm$x$}\|) &\le& W(t,\mb{\bm$x$}) \le \psi_2(\|\mb{\bm$x$}\|),\\
\psi_3\left(\| \mb{\bm$x$}\| \right) &\le& - \dot{W}(t,\mb{\bm$x$})
\end{eqnarray*}
for all $t$ and all $\mb{\bm$x$}$.

If we apply the minmax control strategy
\[
\mb{\bm$u$}(t,\mb{\bm$x$})=\left\{\begin{array}{ccc}
-\frac{\mb{\bm$\mu$}(t,\mb{\bm$x$})}{\|\mb{\bm$\mu$}(t,\mb{\bm$x$})\|} \rho(t,\mb{\bm$x$})&\mb{if}& (t,\mb{\bm$x$}) \notin \Omega\\
\in \left\{\mb{\bm$p$}\in \mb{\Bb R}^m : \|\mb{\bm$p$}\| \le \rho(t,\mb{\bm$x$})\right\} &\mb{if}& (t,\mb{\bm$x$})\in \Omega,
\end{array} \right.
\]
where $\mb{\bm$\mu$}(t,\mb{\bm$x$})=\mb{\bm$G$}^T(t,\mb{\bm$x$}) \nabla_{\mb{\bm$x$}}W(t,\mb{\bm$x$})$, and $\Omega=\left\{(t,\mb{\bm$x$}):\mb{\bm$\mu$}=\mb{\bm$0$}\right\}$, then the equilibrium state $\mb{\bm$x$}=\mb{\bm$0$}$ is a globally uniformly asymptotically stable solution of the closed-loop system. However, the above controller is discontinuous in $\mb{\bm$x$}$. Following Corless and Leitmann~\cite{corless81} we modify it to make it continuous by introducing a boundary layer. The proposed boundary layer controller is
\[
\mb{\bm$u$}(t,\mb{\bm$x$})=\left\{\begin{array}{ccc}
-\frac{\mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$})}{\| \mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$})\|} \rho(t,\mb{\bm$x$}) &\mb{if}& \| \mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$}) \|> \nu\\
\|\mb{\bm$u$}(t,\mb{\bm$x$})\| \le \rho(t,\mb{\bm$x$}) &\mb{if}& \|\mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$})\| \le \nu,
\end{array}\right.
\]
where $\nu$ is the thickness of the boundary layer about the naturally defined switching surface $\left\{(t,\mb{\bm$x$}): \| \mb{\bm$\mu$}(t,\mb{\bm$x$}) \rho(t,\mb{\bm$x$}) \|=0 \right\}$. A particular example of the above type of a controller is
\[
\mb{\bm$u$}(t,\mb{\bm$x$})=\left\{\begin{array}{ccc}
-\frac{\mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$})}{\| \mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$})\|} \rho(t,\mb{\bm$x$}) &\mb{if}& \| \mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$}) \|> \nu\\
-\frac{\mb{\bm$\mu$}(t,\mb{\bm$x$})}{\nu}\rho^2(t,\mb{\bm$x$} ) &\mb{if}& \| \mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$}) \| \le \nu.
\end{array}\right.
\]
We shall now show that the closed-loop system driven by the boundary layer controller is practically stable with respect to a neighborhood of the zero state. This neighborhood can be made arbitrarily small by letting $\nu \to 0$. Before we state the results, we first compute the time derivative of a Lyapunov function candidate on the trajectories of the closed-loop system. We consider the Lyapunov function $W$ we constructed for the uncontrolled nominal system. Then, the corresponding time derivative of $W$ evaluated on the solutions of the closed-loop system is
\begin{eqnarray*}
\dot{W}(t,\mb{\bm$x$}) &=& \frac{\partial W}{\partial t}+\nabla_{\mb{\bm$x$}}^T W \left( \mb{\bm$f$}(t,\mb{\bm$x$})+\mb{\bm$G$}(t,\mb{\bm$x$})(\mb{\bm$u$}+\mb{\bm$\xi$}(t,\mb{\bm$x$})\right)\\
&\le& -\psi_3(\| \mb{\bm$x$} \|)+ \nabla_{\mb{\bm$x$}}^T W \mb{\bm$G$}(t,\mb{\bm$x$}) \mb{\bm$u$} + \| \mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$})\|.
\end{eqnarray*}
We now consider two cases. The first case is when $\|\mb{\bm$\mu$}(t,\mb{\bm$x$})\|>\nu$. Taking into account the controller structure for this case yields
\begin{eqnarray*}
\dot{W}(t,\mb{\bm$x$}) &\le& -\psi_3(\|\mb{\bm$x$}\|) - \nabla_{\mb{\bm$x$}}^T W \mb{\bm$G$}(t,\mb{\bm$x$})\frac{\mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$})}{\| \mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$})\|} \rho(t,\mb{\bm$x$}) + \| \mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$})\|\\
&=& -\psi_3(\|\mb{\bm$x$}\|) - \frac{\|\mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$})\|^2}{\|\mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$})\|} + \|\mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$})\|\\
&=& -\psi_3(\|\mb{\bm$x$}\|).
\end{eqnarray*}
If, on the other hand, $\|\mb{\bm$\mu$}(t,\mb{\bm$x$})\| \le \nu$, then
\begin{eqnarray*}
\dot{W}(t,\mb{\bm$x$}) &\le& -\psi_3(\|\mb{\bm$x$}\|) + \nabla_{\mb{\bm$x$}}^T W \mb{\bm$G$}(t,\mb{\bm$x$})\mb{\bm$u$} + \|\mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$})\|\\
&\le& -\psi_3(\|\mb{\bm$x$}\|) + \|\nabla_{\mb{\bm$x$}}^T W \mb{\bm$G$}(t,\mb{\bm$x$})\|\,\|\mb{\bm$u$}\| + \|\mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$})\|\\
&\le& -\psi_3(\|\mb{\bm$x$}\|) + 2\|\mb{\bm$\mu$}(t,\mb{\bm$x$})\rho(t,\mb{\bm$x$})\|\\
&\le& -\psi_3(\|\mb{\bm$x$}\|) + 2\nu.
\end{eqnarray*}
We conclude that for all $t$ and all $\mb{\bm$x$}$ we have
\[
\dot{W}(t,\mb{\bm$x$}) \le -\psi_3(\|\mb{\bm$x$}\|)+ 2\nu.
\]
To proceed further we need an assumption on the function $\psi_3$.\\
{\bf A1:} If $\psi_3$ is bounded and
\[
\lim_{r \to \infty}\psi_3(r)=l<\infty
\]
then we assume
\[
2\nu < l.
\]
\begin{proposition}
\label{prop:prac_ubd}
The closed-loop system driven by the boundary layer controller is uniformly bounded.
\end{proposition}
{\bf Proof}\quad Let $R$ be such that
\[
\psi_3(R)=2\nu.
\]
Note that $R$ is well defined by virtue of {\bf A1} and the fact that $\psi_3$ is strictly increasing. Since $\psi_3$ is strictly increasing, we can write
\[
R=\psi^{-1}_3(2\nu).
\]
Figure~\ref{fig:psi_the} illustrates the above construction.
\begin{figure}
\vspace{3in}
\hskip1.5in\special{bmp:c:/mybook/psi_the.bmp x=4in y=3in}
\caption{Finding $R$ given $\nu$.}
\label{fig:psi_the}
\end{figure}
Given $d>0$, where $\| \mb{\bm$x$}_0\| \le d$. Let
\[
\tilde{d}=\max \{d,R\}.
\]
We claim that
\[
b=b(d)=\psi^{-1}_1 (\psi_2(\tilde{d})),
\]
that is, the system, or equivalently its null solution, is uniformly bounded. In other words, if $\| \mb{\bm$x$}_0 \| \le d$ then $\|\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)\|\le b$ for $t \ge t_0$. First note that $\|\mb{\bm$x$}_0 \| \le \tilde{d}$ and that $\psi_1(b)=\psi_2(\tilde{d})$. Hence $\tilde{d} \le b$---see Figure~\ref{fig:uni_bdd} for an illustration of this construction.
\begin{figure}
\vspace{3in}
\hskip1.5in\special{bmp:c:/mybook/uni_bdd.bmp x=4in y=3in}
\caption{Finding $b$ given $d$.}
\label{fig:uni_bdd}
\end{figure}
We have
\begin{eqnarray*}
\dot{W}(\mb{\bm$x$}(t)) &\le& -\psi_3 ( \| \mb{\bm$x$}(t) \| )+2\nu\\
&=& -\psi_3 ( \| \mb{\bm$x$}(t) \| )+\psi_3(R).
\end{eqnarray*}
If $\| \mb{\bm$x$}_0 \| > R$, then $\dot{W}<0$ and thus
\begin{eqnarray*}
\psi_1(b) &=& \psi_2(\tilde{d})\\
&\ge& W(t_0;t_0,\mb{\bm$x$}_0)\\
&\ge& W(t;t_0,\mb{\bm$x$}_0)\\
&\ge& \psi_1(\| \mb{\bm$x$}(t)\|),
\end{eqnarray*}
which implies that $\| \mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)\| \le b$ for $t \ge t_0$.

If, on the other hand, $\| \mb{\bm$x$}_0 \| \le R$, then since $\dot{W}(\mb{\bm$x$}(t)) \le -\psi_3 ( \| \mb{\bm$x$}(t) \| )+2\nu$, the solution may reach the sphere $\| \mb{\bm$x$}_0 \| = R$ at some time $t_2 \ge t_0$. However, $\dot{W} < 0$ outside $\| \mb{\bm$x$}_0 \| = R$. Therefore, the solution cannot leave the ball $\| \mb{\bm$x$}_0 \| = b$. Thus the null solution is uniformly bounded.\\
\hspace*{5.5in} $\Box$\\
We illustrate Proposition~\ref{prop:prac_ubd} with Figure~\ref{fig:unib_2d}.
\begin{figure}
\vspace{3in}
\hskip1.5in\special{bmp:c:/mybook/unib_2d.bmp x=4in y=3in}
\caption{A two-dimensional illustration of an uniformly ultimately bounded solution.}
\label{fig:unib_2d}
\end{figure}
The initial condition in Figure~\ref{fig:unib_2d} satisfies $\| \mb{\bm$x$}_0 \|< d$, and $d<R$. The trajectory is confined to stay within the ball of radius $b=\psi_1^{-1}(\psi_2(R))$.

We now show, using arguments similar to those of Corless and Leitmann~\cite{corless81}, that the closed-loop system is uniformly ultimately bounded with respect to the ball of radius $\tilde{b}>\psi^{-1}_1(\psi_2(R))$.
\begin{thm}
If 
\[
\mb{\bm$x$}(t)=\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)
\]
is a solution for $[t_0,\infty)$ with $\| \mb{\bm$x$}_0 \| \le d$, then for a given
\[
\tilde{b} > \psi_1^{-1}(\psi_2(R)),
\]
$\|\mb{\bm$x$}(t;t_0,\mb{\bm$x$}_0)\| \le \tilde{b}$ for all $t \ge t_0 + T(\tilde{b},d)$, where
\[
T(\tilde{b},d)=\left\{\begin{array}{ccc}
0 & \mb{if} & r \le \tilde{R}\\
\frac{\psi_2(d)-\psi_1(\tilde{R})}{\psi_3(\tilde{R})-2\nu}& \mb{if}& r>\tilde{R},
\end{array}\right.
\]
and
\[
\psi_2\left(\tilde{R}\right) = \psi_1(\tilde{b}).
\]
\end{thm}
{\bf Proof}\quad Since by assumption
\[
\psi_1(\tilde{b}) > \psi_2(R),
\]
and
\[
\psi_1(\tilde{b}) = \psi_2(\tilde{R}),
\]
we have
\[
\tilde{R} \ge R.
\]
We consider two cases. The first case is when $d \le \tilde{R}$. Then by uniform boundedness, for all $t \ge t_0$
\[
\| \mb{\bm$x$}(t) \| < b,
\]
for any $b$ such that $\psi_1(b) > \psi_2(\tilde{R})$, that is, $\tilde{b}=b(d)$. Hence in this case $T(\tilde{b},d)=0$.

We now consider the case when $d>\tilde{R}$. We show, by contradiction, that for some $t_1 \in [t_0,t_0+T(\tilde{b},d)]$
\[
\| \mb{\bm$x$}(t_1) \| \le \tilde{R}.
\]
From this we shall conclude the result by referring to the previous case. So suppose that
\[
\| \mb{\bm$x$}(t) \| > \tilde{R}
\]
for all $t \in \left[t_0, t_0 + T(\tilde{b},d) \right]$. Then,
\begin{eqnarray*}
\psi_1\left( \| \mb{\bm$x$}\left(t_0+T(\tilde{b},d)\right);t_0,\mb{\bm$x$}_0)\| \right) &\le& W \left( t_0+T(\tilde{b},d),\mb{\bm$x$}\left(t_0+T(\tilde{b},d)\right);t_0,\mb{\bm$x$}_0)\right)\\
&=& W(t_0,\mb{\bm$x$}_0)+\int_{t_0}^{t_0+T(\tilde{b},d)}\dot{W}dt\\
&=& W(t_0,\mb{\bm$x$}_0)-\int_{t_0}^{t_0+T(\tilde{b},d)}\left(-\dot{W} \right)dt\\
&\le& \psi_2(d) - \int_{t_0}^{t_0+T(\tilde{b},d)}\left(\psi_3(\tilde{R})-2\nu\right)dt.
\end{eqnarray*}
Note that $\psi_3(\tilde{R})-2\nu \ge 0$ because $\tilde{R} \ge R$. We proceed with the evaluation to get
\begin{eqnarray*}
\psi_1\left(\| \mb{\bm$x$}\left(t_0+T(\tilde{b},d)\right);t_0,\mb{\bm$x$}_0)\| \right) &\le& \psi_2(d)-\left(\psi_3(\tilde{R})-2\nu\right)T(\tilde{b},d)\\
&=& \psi_2(d)-\left(\psi_3(\tilde{R})-2\nu\right)\frac{\psi_2(d)-\psi_1(\tilde{R})}{\psi_3(\tilde{R})-2\nu}\\
&=&\psi_1\left(\tilde{R}\right).
\end{eqnarray*}
Thus, since $\psi_1$ is strictly increasing
\[
\| \mb{\bm$x$}\left(t_0+T(\tilde{b},d)\right);t_0,\mb{\bm$x$}_0)\| \le \tilde{R}
\]
which contradicts our assumption that for all $t \in \left[t_0, t_0 + T(\tilde{b},d) \right]$ it should have been $\| \mb{\bm$x$}(t)\| > \tilde{R}$. Hence, there exists $t_1 \in \left[t_0, t_0 + T(\tilde{b},d) \right]$ such that $\| \mb{\bm$x$}(t_1)\| \le \tilde{R}$. Then using the same argument as in the previous case we can show that
\[
\| \mb{\bm$x$}(t) \| < \tilde{b}
\]
for all $t \ge t_1$ and the proof is complete.\\
\hspace*{5.5in} $\Box$

We have the following consequence of the above theorem.
\begin{corollary}
The ball
\[
\left\{ \mb{\bm$x$}:\|\mb{\bm$x$}\| \le \psi_1^{-1}(\psi_2(R)) \right\}
\]
is uniformly stable.
\end{corollary}
{\bf Proof}\quad Let
\[
\varepsilon > \psi_1^{-1}(\psi_2(R)),
\]
and consider $\delta=\delta(\varepsilon)$ obtained from
\[
\psi_1(\varepsilon)=\psi_2(\delta).
\]
Obviously $\delta \ge R$---see Figure~\ref{fig:ub_ball}.
\begin{figure}
\vspace{3in}
\hskip1.5in\special{bmp:c:/mybook/ub_ball.bmp x=4in y=3in}
\caption{Constructing $\delta$ given $\varepsilon$ for an uniformly stable ball.}
\label{fig:ub_ball}
\end{figure}
Hence by uniform boundedness if $\| \mb{\bm$x$}_0 \| \le \delta$, then $\| \mb{\bm$x$}(t) \| \le \varepsilon$ for all $t \ge t_0$ which means that the ball is uniformly stable.\\
\hspace*{5.5in} $\Box$

\section{Exercises}
\begin{exercise}\rm
Show
\end{exercise}