\documentclass[12pt]{article}

\usepackage{amsmath,amssymb,amsthm}

\usepackage{color}

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

\usepackage{tikz}
\usepackage{pgfplots}
\pgfplotsset{compat=1.16}
%\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[shortlabels]{enumitem}
\setlist[enumerate]{leftmargin=*,label=(\alph*)}

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

\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

\allowdisplaybreaks

%\pagestyle{empty}

\begin{document}
\centerline{Math 322 worksheet for Wednesday, Week 7}
\bigskip

\textbf{Definition.} An \emph{equilibrium point} for a system of
differential equations in~$\R^n$
\begin{align*}
  x'&= f(x)
\end{align*}
is a point~$p\in\R^n$ such that~$f(p)=0$.
\medskip

The reason for the terminology is that if~$p$ is an equilibrium point then a
solution ({\em the solution} if~$f$ is continuously differentiable) with initial
condition~$x(0)=p$ is the constant solution~$x(t)=p$.

We hope to get a qualitative sense of the solutions to our system near an
equilibrium point~$p$ by replacing the system with a linear approximation:
\begin{align*}
  x'&= Jf_p
\end{align*}
where~$Jf_p$ is the Jacobian matrix for~$f$ at~$p$.
\medskip

Consider the system of equations
\begin{align*}
  x'&= (x^2-1)y\\
  y'&= (1-y^2)\left( x+\frac{3}{10}y \right).
\end{align*}
So in this case~$f(x,y)=((x^2-1)y,(1-y^2)\left( x+\frac{3}{10}y \right))$.
\medskip

\textbf{Problem 1.} Find all equilibrium points for the system and plot them in the
plane.
\bigskip

\textbf{Problem 2.} Compute the Jacobian matrix~$Jf_{(x,y)}$ for our~$f$ at an
arbitrary point~$(x,y)$.
\bigskip

\textbf{Problem 3} For each equilibrium point~$p$, analyze the linear system
\[
  \left(\begin{array}{c}
      x'\\y'
  \end{array} \right)
  =
Jf_p
  \left(\begin{array}{c}
      x\\y
  \end{array} \right)
\]
by looking at the eigenvalues of~$Jf_p$.  Do you get a saddle? A stable
focus or node? An unstable focus or node? A center? (See the last page for a
quick guide.)
\bigskip

\textbf{Problem 4.} What does the vector field look like along the line~$x=1$ and
along the line~$x=-1$?  What can you say about the special behavior of solutions with
an initial condition~$(\pm 1,y_0)$?  Interpret this geometrically.
\bigskip

\textbf{Problem 5.} What does the vector field look like along the line~$y=1$ and
along the line~$y=-1$?  What can you say about the special behavior of solutions with
an initial condition~$(x_0,\pm 1)$?  Interpret this geometrically.
\newpage

\centerline{\sc equilibrium points for linear systems in~$\R^2$}
\medskip

Let~$A\in M_2(\R)$.  Let~$\tau$ be the trace of~$A$, and let~$\delta$ be the
determinant of~$A$. The characteristic polynomial for~$A$ will factor
as
\begin{align*}
p(x)
&= (\lambda_1-x)(\lambda_2-x)\\
&= x^2-(\lambda_1+\lambda_2)x+\lambda_1\lambda_2\\
&= x^2-\tau x+\delta
\end{align*}
where~$\lambda_1$ and~$\lambda_2$ are the eigenvalues of~$A$.
Setting~$p(x)=0$ and solving gives an alternate description of the eigenvalues:
\[
  x = \frac{\tau\pm\sqrt{\tau^2-4\delta}}{2}.
\]
If~$\delta=0$, then at least one of the eigenvalues is zero, and we have a {\bf degenerate system}.

\fbox{$\delta=0$}\ \  {\bf degenerate}.

\fbox{$\delta<0$}\ \  real eigenvalues, opposite signs $\Rightarrow$ {\bf saddle}.

\fbox{$\delta>0$,~$\tau^2-4\delta\geq0$}\ \ real eigenvectors, same
signs $\Rightarrow$  node.

\rule{0pt}{0pt}\hspace{12pt}$\tau<0$ $\Rightarrow$ {\bf stable node}\newline
\rule{0pt}{0pt}\hspace{12pt}$\tau>0$ $\Rightarrow$ {\bf unstable node}.

\fbox{$\delta>0$,~$\tau^2-4\delta<0$}\ \ nonreal eigenvectors $\Rightarrow$ swirling vector field.

\rule{0pt}{0pt}\hspace{12pt}$\tau<0$ $\Rightarrow$ {\bf stable focus}\newline
\rule{0pt}{0pt}\hspace{12pt}$\tau>0$ $\Rightarrow$ {\bf unstable focus}\newline
\rule{0pt}{0pt}\hspace{12pt} $\tau=0$ $\Rightarrow$ {\bf center}.


\begin{center}
  \begin{tikzpicture}[scale=0.8]
    \draw (-5,0)--(5,0);
    \draw (0,0)--(0,5);
    \node at (5.3,0) {$\tau$};
    \node at (0,5.4) {$\delta$};
    \node at (1.25,3.0) {unstable};
    \node at (1.25,2.5) {focus};
    \node at (-1.25,3.0) {stable};
    \node at (-1.25,2.5) {focus};
    \node at (-3.75,1.8) {stable};
    \node at (-3.75,1.3) {node};
    \node at (3.75,1.8) {unstable};
    \node at (3.75,1.3) {node};
    \node at (0,-1) {saddle};
    \node at (0,0) {degenerate};
    \node at (-3,0) {degenerate};
    \node at (3,0) {degenerate};
    \node[rotate=90] at (-0.04,1.2) {center};
    \begin{scope}[xshift=-3.43cm, yshift=-4.6mm]
  \begin{axis}[ 
      %title style={at={(0.5,0)},anchor=north,yshift=-30},
      %title={Graph of~$f(x)=x^2$ and its tangent line at~$x=3$.},
      axis lines*=middle,
      axis equal image,
      ticks=none,
      xmin=-5,
      xmax=5,
      %ymin=-0.2,
      ymax=6.5
    ]
    %\fill[blue] (axis cs: 1,1) circle (2pt);  
    %\fill[draw=black,fill=white] (axis cs:1,0) circle (2pt);  
    \addplot[domain=-5:5,thick, blue] {x*x/4};
  \end{axis}
  \end{scope}
  \end{tikzpicture}
\end{center}
\end{document}


