\documentclass[12pt]{article}

\usepackage{amsmath}
\usepackage{amssymb}

\usepackage{xcolor}

%\usepackage[textheight=8.5in, margin=1.6in]{geometry}
\usepackage[total={6in,8in}]{geometry}

\usepackage{tikz}
%\usepackage{pgfplots}
%\tikzstyle{ball} = [circle,shading=ball, ball color=black,
%    minimum size=1mm,inner sep=1.3pt]
%\usetikzlibrary{arrows.meta}
%\tikzset{>={Latex[length=2mm,width=1.5mm]}}

\usepackage{graphicx}

\usepackage{needspace}  % <-- to prevent bad pagebreak over environment titles
% Use:  \Needspace{3\baselineskip} before opening an environment

\usepackage{enumitem}
\setlist[enumerate]{leftmargin=*}

%\usepackage[colorlinks=true,citecolor=black,linkcolor=black,urlcolor=blue]{hyperref}

\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\sn}{\mathfrak{S}}
\newcommand{\ve}{\varepsilon}
\newcommand{\ips}[2]{\langle{#1}, {#2}\rangle}
\DeclareMathOperator{\tr}{tr}
\DeclareMathOperator{\Span}{\mathrm{Span}}
\DeclareMathOperator{\rk}{\mathrm{rank}}
\DeclareMathOperator{\trace}{\mathrm{tr}}
\DeclareMathOperator{\im}{\mathrm{im}}
\DeclareMathOperator{\Sym}{\mathrm{Sym}}
\DeclareMathOperator{\Hom}{\mathrm{Hom}}
\DeclareMathOperator{\diag}{\mathrm{diag}}
\DeclareMathOperator{\glb}{\mathrm{glb}}
\DeclareMathOperator{\lub}{\mathrm{lub}}

\parskip = 0.05in
\parindent = 0.0in

%\pagestyle{empty}

\begin{document}
\centerline{Math 322 lecture for Friday, Week 9}
\bigskip

\centerline{\sc liapunov functions}
\medskip

\textbf{Theorem.}  Let~$f\in C^1(E)$ and~$f(x_0)=0$.  Let~$V\colon E\to\R$ also
be~$C^1$ (continuously differentiable).  Suppose that~$V(p)\geq 0$ and~$V(p)=0$
if and only if~$p=x_0$. Then:
\begin{enumerate}
  \item If~$\dot{V}$ is negative semidefinite ($\dot{V}(p)\leq0$ for
    all~$p\in E\setminus\left\{ x_0 \right\}$)  then~$x_0$ is stable.
  \item If~$\dot{V}$ is negative definite ($\dot{V}(p)<0$ for all~$p\in
    E\setminus\left\{ x_0 \right\}$)
    then~$x_0$ is asymptotically stable.
  \item If~$\dot{V}$ is positive definite ($\dot{V}(p)>0$ for all~$p\in
    E\setminus\left\{ x_0 \right\}$), then~$x_0$ is unstable.
\end{enumerate}
\smallskip

{\bf Proof.} As before, we may assume~$x_0=(0,0)$ is the equilibrium point.  Part (1) was proved in the last lecture.  

(2) Last time, we were in the midst of proving part~(2).  Using the notation
from last time, so far, we have shown that for every sequence~$t_{1}<t_{2}<\cdots$
such that $\lim_{n \to \infty} t_{n}=\infty$, there exists a
subsequence~$\{t_{n_{k}}\}$ such that~$\lim_{k \to \infty}
\phi(t_{n_{k}},p)=0\in\R^{n}$.

We now need to show~$\lim_{t\to\infty}\phi(t,p)=x_0=0$.  If not, there exists
an~$\eta>0$ such that for all~$n$, there exists~$t_n>n$ such that
\begin{equation}
  |\phi(t_n,p)|\geq\eta>0.
  \label{eqn: 09wed1}
\end{equation}
We may assume that the sequence~${t_n}$ is increasing.  However, by
Bolzano-Weierstrass, there again exists a subsequence~$\left\{ t_{n_k}
\right\}$ of~$\left\{ t_n \right\}$ such that~$\phi(t_{n_k},p)$ converges, and
as we have seen, it must converge to~$0$.  But that's impossible in light
of~\eqref{eqn: 09wed1}.

(3) Finally, now suppose that~$\dot{V}(p)>0$ for all~$p\in E\setminus\left\{ x_0 \right\}$.  
Choose~$\ve>0$ such that~$\overline{B_{\ve}(0)}\subset E$.  We'll show
  that given any point~$p\in B_{\varepsilon}(0)\setminus\{0\}$, we have that~$\phi_t(p)$
  leaves~$B_{\ve}(0)$ at some point, i.e., there exists~$t\geq 0$
  such that~$|\phi_t(p)|\geq\ve$. 

  Given~$p\in B_{\varepsilon}(0)\setminus\{0\}$, since~$V$ is strictly increasing on trajectories,
  \[
    V(\phi_t(p))>V(\phi_0(p))=V(p)>0
  \]
  for all~$t>0$.  Thus,~$\phi_t(p)$ is bounded away from~$0$.
  Say~$|\phi_t(p)|\geq\eta>0$ for all~$t\geq0$.
  Since~$\eta\leq|\phi_{0}(p)|=|p|<\varepsilon$, it follows
  that~$\eta<\varepsilon$. Define
  \[
    m := \min_{y:\eta\leq|y|\leq\ve}\dot{V}(y),
  \]
  which exists since~$\dot{V}$ is continuous and~$y$ is restricted to a compact
  set. In fact, for that same reason,~$m=V(q)$ for some point in the set over
  which we are minimized.  Therefore,~$m>0$.  Supposing for contradiction that~$\phi_t(p)$ stays
  inside~$B_{\ve}(x_0)$ for all~$t\geq0$, we have~$\dot{V}(\phi_t(p)\geq m$ for
    all~$t\geq0$.  Hence,
    \[
      V(\phi_t(p))-V(p)=V(\phi_t(p))-V(\phi_0(p))=\int_{s=0}^t\dot{V}(\phi_s(p))\,ds\geq mt\to\infty
    \]
    as~$t\to\infty$.  But since~$V$ is continuous, it achieves a maximum
    on~$\overline{B_{\ve}(x_0)}$---a contradiction.
\hfill$\Box$
\medskip

{\bf Example.}  Consider the system
\begin{align*}
  x'&= -2y+yz\\
  y'&= x-xz\\
  z'&= xy.
\end{align*}
and the Jacobian at the origin is
\[
  J(0)=
  \left(\begin{array}{ccc}
      0&-2&0\\
      1&0&0\\
      0&0&0
  \end{array} \right).
\]
The characteristic polynomial is
\[
\det
  \left(\begin{array}{ccc}
      -x&-2&0\\
      1&-x&0\\
      0&0&-x
  \end{array} \right)
  =-x^3-2x=-x(x^2+2).
\]
So the eigenvalues are~$0,\pm\sqrt{2}i$.  So the origin is a nonhyperbolic
equilibrium point.  To determine stability, we look for a suitable Liapunov function.
We guess a function of the form
\[
  V = ax^2+by^2+cz^2
\]
with positive constants~$a,b,c$. We have
\begin{align*}
  \dot{V}&= 2axx'+2byy'+2czz'\\
  &= 2ax(-2y+yz)+2by(x-xz)+2cz(xy)\\
  &= 2(-2a+b)xy+(a-b+c)xyz.
\end{align*}
Take~$a=c=1$ and~$b=2$, and we get~$V=x^2+2y^2+z^2$ with~$\dot{V}=0$.  This
means that trajectories stay on the ellipsoids that are level sets of~$V$.
\end{document}


