\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\setlength{\textwidth}{6.5in} \setlength{\textheight}{9in}
\setlength{\topmargin}{-.25in} \setlength{\oddsidemargin}{-.25in}
\newcommand\br{\textbf{R}}
\pagestyle{empty}
\begin{document}

\parindent=0pt

\bf{Question}

Sketch the \emph{singular set} $\Sigma$ and the
\emph{discriminant} $\Delta=F(\Sigma)$ for each of the following
maps $F:\mathbf{R}^2\rightarrow\mathbf{R}^2$:

\begin{description}

\item[(i)]
$F(x_1,x_2)=(x_1,x_1^2+x_2^2)$

\item[(ii)]
$F(x_1,x_2)=(x_1,x_1^3-x_1+x_2^2)$

\item[(iii)]
$F(x_1,x_2)=(x_1^2-x_2^2,2x_1x_2)$

\item[(iv)]
$F(x_1,x_2)=(x_1,x_2^4+x_1x_2^2)$.

\end{description}

In each case label the various connected regions of the complement
of $\Delta$ in $\mathbf{R}^2$ according to the number of points of
$F^{-1}(q)$ for $q$ in each region.




\bf{Answer}

\begin{description}
\item{(i)}
$DF(x_1, x_2) = \left ( \begin{array}{cc} 1 & 0 \\ 2x_1 & 2x_2
\end{array} \right )$.

det$=2x_2$, $\Rightarrow \Sigma$ is: $x_2=0$.

$F(x_1,0)=(x_1, x_1^2)$ $\Rightarrow \Delta$ is: $v=u^2$.

\begin{center}
\epsfig{file=342-6A-1.eps, width=70mm}
\end{center}

\item{(ii)}
$DF(x_1, x_2) = \left ( \begin{array}{cc} 1 & 0 \\ 3x_1^2-1 & 2x_2
\end{array} \right )$.

det$=2x_2$, $\Rightarrow \Sigma$ is: $x_2=0$.

$F(x_1,0)=(x_1, x_1^3 - 1)$ $\Rightarrow \Delta$ is: $v=u^3 - u$.

\begin{center}
\epsfig{file=342-6A-2.eps, width=70mm}
\end{center}

\item{(iii)}
$DF(x_1, x_2) = \left ( \begin{array}{cc} 2x_1 & -2x_2 \\ 2x_2 &
2x_1
\end{array} \right )$.

det$=4(x_1^2 + x_2^2$, $\Rightarrow \Sigma$ is $(0,0)$ only.

$\Delta = \{ F(0,0) \} = (0,0)$ only.

In polar coordinates $(x_1, x_2) = r(\cos\theta, \sin\theta)$ we
see $$(u,v) = r^2)\cos 2\theta, \sin 2\theta)$$ i.e. $F(x)=x^2$ in
complex notation.

\begin{center}
\epsfig{file=342-6A-3.eps, width=70mm}

Angles are doubled, radii are squared.
\end{center}

\item{(iv)}
$DF(x_1, x_2) = \left ( \begin{array}{cc} 1 & 0 \\ x_2^2 & 4x_2^3
+ 2x_1x_2
\end{array} \right )$.

$\det=4x_2^3+2x_1x_2$, $\Rightarrow \Sigma$ is: $x_2=0$ and
$x_1=-2x_2^2$.

$F(x_1,0) = (x_1,0)$, $F(-2x_2^2, x_2) = (-2x_2^2, -x_2^4)$,
giving part of the parabola $\ds v=-\left ( \frac{u}{2} \right
)^2$ with $u \le 0$.

\begin{center}
\epsfig{file=342-6A-4.eps, width=70mm}
\end{center}
\end{description}


\end{document}