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\begin{document}
\centerline{Math 322 lecture for Wednesday, Week 10}
\bigskip

\centerline{\sc global theory for nonlinear systems: index theory}
\medskip

A {\em Jordan curve}~$C$ is the injective continuous image~$\gamma\colon
S^1\to\R^2$ of a circle into the plane.  Equivalently, it is the continuous
image of an interval~$\gamma\colon [0,1]\to\R^2$ that is injective on~$[0,1)$
and such that~$\gamma(0)=\gamma(1)$.  The Jordan curve theorem (first
conjectured by Bolzano) states that a Jordon curve divides the plane into two
connected components.  We will impose the further condition that~$\gamma$ be
piecewise smooth (continuous derivatives except at a finite number of points). 

Let~$f(x,y)=(P(x,y),Q(x,y))$ be a smooth vector field in the plane, and
let~$C$ be a Jordan curve.  A {\em critical point} for~$f$ is a
point~$(x_0,y_0)$ where~$f(x_0,y_0)=0$.  (Thus, a critical point would be an
  equilibrium point for the corresponding system of differential equations.)
\medskip

\textbf{Definition.} The index~$I_f(C)$ of~$C$ relative to~$f$ is
\[
  I_f(C):=\frac{\Delta\theta}{2\pi}
\]
where~$\Delta\theta$ is the change in angle of~$f(x,y)$ as~$(x,y)$ travels
around~$C$ counterclockwise.  
\medskip

{\bf Exercises.}
\begin{enumerate}
  \item 
For each of the following vector fields, (i) draw the flow near
the origin and draw a circle~$C$ containing the origin; (ii) pick some points
on~$C$, and draw each point as a vector (with tail at the origin) on a separate
picture of~$\R^2$; (iii) compute the index:
\[
\begin{array}{ll}
  \text{(a)}\quad f(x,y)=(-1,-1)\qquad&\text{(b)}\quad f(x,y)=(-x,-y)\\[5pt]
  \text{(c)}\quad f(x,y)=(-y,x)\qquad&\text{(d)}\quad f(x,y)=(-x,y).
\end{array}
\]
\item How does the index change in (a)--(d) if~$f$ is replaced by~$-f$?
\item How would the index change if~$C$ were replaced by an ellipse?
\end{enumerate}
\medskip

{\bf Calculation of the index.} Let~$\gamma(t)=(x(t),y(t))$ be a parametrization
of~$C$.  By translating, if necessary, we may assume the origin is in the
interior of~$C$.  Consider the composition of mappings
\[
  [0,1)\xrightarrow{\gamma}\R^2\xrightarrow{f}\R^2.
\]
We are interested in the change in the angle of~$f\circ\gamma$ as~$t$ goes
from~$0$ to~$1$.  Write~$f\circ\gamma$ in polar coordinates:
\[
  P(x,y)= r\cos(\theta)\quad~Q(x,y)=r\sin(\theta)
\]
where~$x,y,r,\theta$ are functions of~$t$.  Then
\begin{align*}
  P'&= r'\cos(\theta)-r\theta'\sin(\theta)\\
  Q'&= r'\sin(\theta)+r\theta'\cos(\theta),
\end{align*}
and it's easy to check that
\[
  r^2\theta'=PQ'-QP'
\]
where~$r^2=P^2+Q^2$.  Therefore, the change in angle is
\[
  \Delta\theta=\int_{t=0}^{1}\frac{PQ'-QP'}{P^2+Q^2}\,dt
  =\int_{t=0}^1(P,Q)\cdot\left( \frac{Q'}{P^2+Q^2},-\frac{P'}{P^2+Q^2}
  \right)\,dt
  =\oint_{C}\frac{PdQ-QdP}{P^2+Q^2}.
\]
So the index is
\[
  I_f(C)=\frac{\Delta\theta}{2\pi}=\frac{1}{2\pi}\oint_{C}\frac{PdQ-QdP}{P^2+Q^2}.
\]
To convert to the language of differential forms, let
\[
  \omega:=\frac{x\,dy-y\,dx}{x^2+y^2}
\]
be the ``flow form'' for the circular vector field~$\left(-
\frac{y}{x^2+y^2},\frac{x}{x^2+y^2} \right)$ on~$\R^2$.  Then the
index is calculated by integrating the pullback of~$\omega$ along~$f$
over~$C$:
\[
  I_f(C)=\frac{1}{2\pi}\oint_{\gamma}f^*\omega.
\]
\medskip

{\bf Example.} Let~$f(x,y)=(-y,x)$, and let~$C$ be the unit circle centered at
the origin parametrized by~$\gamma(t)=(\cos(t),\sin(t))$ for~$t\in[0,2\pi]$.  Then
\[
  f(\gamma(t))=(-\sin(t),\cos(t)).
\]
\begin{align*}
I_f(C)&=\frac{\Delta\theta}{2\pi}=\frac{1}{2\pi}\oint_{C}\frac{PdQ-QdP}{P^2+Q^2}
\\[10pt]
  &= \frac{1}{2\pi}\int_{C}(P,Q)\cdot\left(
  \frac{Q'}{P^2+Q^2},-\frac{P'}{P^2+Q^2}\right)\,dt\\[10pt]
  &= \frac{1}{2\pi}\int_{t=0}^{1}(-\sin(t),\cos(t))\cdot\left(
  \frac{-\sin(t)}{(-\sin(t))^2+\cos(t)^2},-\frac{-\cos(t)}{(-\sin(t))^2+\cos(t)^2}\right)\,dt\\[10pt]
  &= \frac{1}{2\pi}\int_{t=0}^{2\pi}\,dt= 1.
\end{align*}
\medskip

\textbf{Theorem.} If there are no critical points on~$C$ or in its
interior, then~$I_f(C)=0$.
\begin{proof}
 {\bf Step 1.}  Suppose~$C=C_1+C_2$ as shown below:
 \begin{center}
   \begin{tikzpicture}
     \draw[blue] (0,0) arc (-90:90:1.4cm and 1cm);
     \draw[red] (0,2cm) arc (90:270:1.4cm and 1cm);
     \draw (0,0) to [bend right=3] (0,2);
     \node at (-0.6,1) {$C_1$};
     \node at (0.6,1) {$C_2$};
   \end{tikzpicture}
 \end{center}
 The curves~$C_1$ and~$C_2$ share the vertical middle line.  In the calculating
 of the sum of the indices of~$f$ relative to~$C_1$ and to~$C_2$, the
 contribution from the middle line cancels (imaging traveling along
 both~$C_1$ and~$C_2$ in the counterclockwise direction).  Thus,
 $I_f(C)=I_f(C_{1})+I_{f}(C_{2})$.
 
 {\bf Step 2.}  Next, divide~$C$ into a sum of lots of tiny closed curves:
 \begin{center}
   \begin{tikzpicture}
     \clip (0,0) ellipse (2cm and 1cm);
     \draw (0,0) ellipse (2cm and 1cm);
     \draw[step=0.3,gray,very thin] (-2,-2) grid (6,6);
   \end{tikzpicture}
 \end{center}
 So~$C=C_1+\dots+C_n$.  It suffices to show that~$I_f(C_i)=0$ for all~$i$.  Some
 details and a reference are provided below, but the main idea is that
 since~$f\neq \vec{0}$ on or inside $C$, by taking~$C_i$ small enough, the
 vector field's angle cannot change much along~$C_i$:
 \begin{center}
   \begin{tikzpicture}[scale=0.8]
     \foreach \x in {0,...,10}
     {
       \draw[->] (0,0.3*\x)--(3,3+0.3*\x);
       \draw[blue] (1.5,2.8) rectangle (1.8,3.1);
   }
   \end{tikzpicture}
 \end{center}
 Let~$X$ denote~$C$ union its interior.  Since~$X$ is compact, and the component
 functions~$P$ and~$Q$ of the vector field are continuous, it follows that~$P$
 and $Q$ are uniformly continuous.  That means that given any~$\varepsilon>0$,
 we can make the widths of the~$C_i$ simultaneously small enough to that~$P$
 and~$Q$ change by a value less then~$\varepsilon$ on each~$C_i$.  Also, since~$X$ is compact
 and $f$ is continuous and nonzero in $X$, the value of $|f(x,y)|$ attains a
 nonzero minimum on~$X$. This means that it is possible to take the widths of
 the~$C_{i}$ simultaneously small enough so that the angle of~$f$ on~$X$ varies
 by only a small amount (less than~$2\pi$ is sufficient).  Some details appear
 in our text, Problem~2, Chapter~3.
 
 The result then follows from Step~1: $I_{f}(C)=\sum_iI_{f}(C_{i})=\sum_{i}0=0$.
\end{proof}

\textbf{Corollary.} Let~$C$ be a Jordan curve.  Suppose there are no
critical points on~$C$ but that there may be critical points in its interior.  Let~$C'$ a
Jordan curve in the interior of~$C$, and suppose there are no critical points
on~$C'$, and there are no critical points in the region between~$C$
and~$C'$.  Then~$I_f(C)=I_f(C')$.
\begin{proof}
 Referring to the diagram below, let~$\partial A$ and~$\partial B$ be
 the  Jordan curves forming the boundaries of the closed regions labeled~$A$
 and~$B$.  So both~$\partial A$ and~$\partial B$ are the boundaries of deformed
 rectangles.  Imagine traveling counterclockwise along these curve, computing
 their indices.  You should see that
 \[
   I_f(\partial A+\partial B) = I_f(C)-I_f(C').
 \]
 However, since there are no critical points in~$A$ and none in~$B$, using the
 previous theorem, we have
 \[
   I_f(\partial A+\partial B)=I_f(\partial A)+I_f(\partial B)=0.
 \]
 The result follows.
 
 \begin{center}
   \begin{tikzpicture}[scale=0.7]
     \draw (0,0) circle (3cm);
     \draw (0,0) circle (0.8cm);
     \draw (0.8,0)--(3,0);
     \draw (-0.8,0)--(-3,0);
     \node at (0,1.7) {$A$};
     \node at (0,-1.7) {$B$};
     \node at (-2.4,2.4) {$C$};
     \node at (-0.9,0.7) {$C'$};
   \end{tikzpicture}
 \end{center}
\end{proof}

\textbf{Corollary.} If~$C$ and~$C'$ are Jordan curves containing the same
finite set of critical points in their interiors, then~$I_f(C)=I_f(C')$.
\begin{proof}
 Let~$D$ be a Jordan curve containing all the critical points and
 contained in the interiors of both~$C$ and~$C'$.  Then by the previous
 corollary,
 \[
   I_f(C)=I_f(D)=I_f(C').
 \]
\end{proof}

\textbf{Definition.} Let~$p$ be an isolated critical point of~$f$.
Define the {\em index of~$x$ relative to~$f$} to be
\[
  I_f(p):=I_f(C)
\]
where~$C$ is any Jordan curve containing~$p$ as its only interior critical
point.  (This is well-defined from the previous corollary.)
\medskip

\textbf{Theorem.} Let~$p_1,\dots,p_n$ be the critical points
inside~$C$.  Then
\[
  I_f(C)=\sum_{i=1}^nI_f(p_i).
\]
\smallskip

\Needspace{3\baselineskip}
\begin{proof}
 The proof is similar to that of our first corollary:
 \begin{center}
   \begin{tikzpicture}[scale=0.8]
     \node at (-3,2) {$C$};
     \draw (0,0) ellipse (4cm and 2cm);
     \draw (-2.7,0) circle (0.4 cm);
     \node at (-2.7,0) {$\,p_1$};
     \draw (-4,0)--(-3.1,0);
     \draw (-1.5,0) circle (0.4 cm);
     \node at (-1.5,0) {$\,p_2$};
     \draw (-2.3,0)--(-1.9,0);
     \draw (-0.3,0) circle (0.4 cm);
     \node at (-0.3,0) {$\,p_3$};
     \draw (-1.1,0)--(-0.7,0);
     \draw (0.1,0)--(0.4,0);
     \draw[densely dashed] (0.42,0)--(1.8,0);
     \draw (1.82,0)--(2.3,0);
     \draw (2.7,0) circle (0.4 cm);
     \node at (2.7,0) {$\,p_n$};
     \draw (3.1,0)--(4,0);
     \node at (0,1.1) {$A$};
     \node at (0,-1.1) {$B$};
   \end{tikzpicture}
 \end{center}
 We have
 \[
   0=I_f(\partial A+\partial B)=I_f(C)-\sum_{i=1}^nI_f(p_i).
 \]
\end{proof}
\end{document}
