\documentclass[12pt]{article}
\usepackage{fullpage,amsmath,amsfonts,amsthm,graphicx,mathtools}

\def\d{{\mathrm d}}
\def\e{{\mathrm e}}
\def\R{{\mathbb R}}
\def\v{\boldsymbol{v}}

\newtheorem*{inft}{Inverse Function Theorem}
\newtheorem*{imft}{Implicit Function Theorem}

\begin{document}

\title{Advanced Calculus and \\Methods of Mathematical Physics}
\author{Homework 5}
\date{Due in class Tuesday, March 10, 2020}
\maketitle

\emph{Note:} Assignments marked (*) will not be graded.  Do not turn
them in.  However, they will be discussed in the tutorial and example
solutions will be made available.

\begin{inft}
For $E \subset \R^n$ open and $f \in C^1(E,\R^n)$, suppose that
$Df(a)$ is invertible for some $a \in E$.\footnote{Here, $DF$ denotes
the matrix representation of the derivative $\d f$.  Usually, we do
not make this distinction explicit, but in this context, the focus is
on the block matrix structure so that, following convention, the
notation $DF$, $D_x f$, and $D_y f$ is preferred.}  Then
\begin{itemize}
\item[(i)] There exist open sets $U,V \subset \R^n$ with $a \in U$
such that $f \colon U \to V$ is bijective,
\item[(ii)] $f^{-1} \in C^1 (V,U)$ and
\begin{equation}
  \tag{*}
  Df^{-1} (y) = Df(x)^{-1} 
\end{equation}
with $y = f(x)$.
\end{itemize}
\end{inft}

\begin{imft}
Let $X \subset \R^n$ and
$Y \in \R^m$ be open and $f \in C^1(X \times Y, \R^n)$.  Suppose there
exist $a \in X$ and $b \in Y$ such that $f(a,b)=0$ and $D_xf(a,b)$ is
invertible.  Then there exist an open neighborhood $A$ of $a$, an open
neighborhood $B$ of $b$, and a function $g \in C^1(B,A)$ such that
\[
  f(g(y),y) = 0
\]
for every $y \in B$, and 
\[
  Dg(y) = - D_xf(g(y),y)^{-1} \, D_y f(g(y),y) \,.
\]
\end{imft}

\begin{enumerate}

\item (From Rudin, Exercise 9.17.) Let $f \colon \R^2 \to \R^2$ be
defined by
\[
  f(x) =
  \begin{pmatrix}
    \e^{x_1} \, \cos x_2 \\ \e^{x_1} \, \sin x_2
  \end{pmatrix} \,.
\]
\begin{enumerate}
\item What is the range of $f$?
\item Show that the \emph{Jacobian determinant}, $\operatorname{det}
Df(x)$, is non-zero for every $x \in \R^2$.  Thus every point in
$\R^2$ has a neighborhood in which $f$ is one-to-one.  Nevertheless,
$f$ is not one-to-one on $\R^2$.

\item Put $a = (0, \pi/3)$ and $b = f(a)$.  Find an explicit
formula for $f^{-1}$, compute $Df(a)$ and $Df^{-1}(b)$, and verify
formula (*) above.

\item What are the images under $f$ of lines parallel to the
coordinate axes?

\end{enumerate}

\item *(From the \emph{inverse} to the \emph{implicit function
theorem}.)  Let $X \subset \R^n$ and $Y \in \R^m$ be open and $f \in
C^1(X \times Y, \R^n)$.  Define $F \colon X \times Y \to \R^n \times
\R^m$ as
\[
  F(x,y)
  =
  \begin{pmatrix}
    f(x,y) \\ y
  \end{pmatrix} \,.
\]
\begin{enumerate}
\item Show that $DF$ has the block-matrix structure
\[
  DF =
  \begin{pmatrix}
    D_x f & D_y f \\
    0 & I_m
  \end{pmatrix} \,,
\]
where $I_m$ is the $m \times m$-identity matrix and $D_xf$ and $D_yf$
denote the generalized partial derivatives
\[
  D_x f =
  \begin{pmatrix}
    \partial_{x_1} f_1 & \dots & \partial_{x_n} f_1 \\
    \vdots & & \vdots \\
    \partial_{x_1} f_n & \dots & \partial_{x_n} f_n
  \end{pmatrix}
  \quad \text{and} \quad
  D_y f =
  \begin{pmatrix}
    \partial_{y_1} f_1 & \dots & \partial_{y_m} f_1 \\
    \vdots & & \vdots \\
    \partial_{y_1} f_n & \dots & \partial_{y_m} f_n
  \end{pmatrix} \,.
\]

\item Conclude that $DF$ is invertible if and only if $D_xf$ is
invertible.

\item Suppose that there exist $a \in X$ and $b \in Y$ such that
$f(a,b)=0$ and $D_xf(a,b)$ is invertible.  Argue that there exists an
open neighborhood $V$ of $(0,b)$ and an open neighborhood $U$ of
$(a,b)$ such that $F \colon U \to V$ is bijective with $F^{-1} \in
C^1(V,U)$.

\item Conclude that there exist an open neighborhood $A$ of $a$,
an open neighborhood $B$ of $b$, and a function $g \colon B \to A$
such that
\[
  f(g(y),y) = 0
\]
for every $y \in B$.

\item Finally, argue that $g \in C^1(B,A)$ with
\[
  Dg(y) = - D_xf(g(y),y)^{-1} \, D_y f(g(y),y) \,.
\]

\end{enumerate}

\item *(Kantorovitz, p.~106, Exercise 1.) Show that the equation
\[
  x^5 + y^5 + z^5 = 2 + xyz
\]
determines in a neighborhood of the point $(1,1,1)$ a unique function
$z=z(x,y)$ of class $C^1$, and calculate its partial derivatives with
respect to $x$ and $y$ at the point $(1,1)$.

\item Consider the equation
\[
  \sqrt{x^2 + y^2 + 2 \, z^2} = \cos z
\]
near $(0,1,0)$.  Can you solve for $x$ in terms of $y$ and $z$?  For
$z$ in terms of $x$ and $y$?


\item Show that if $r$ is a simple root of the polynomial
\[
  p(x) = a_0 + a_1 \, x + \dots + a_n \, x^n \,,
\]
then $r$ is a $C^1$ function of the coefficients
$a_0, \dots, a_n$.

\item *Argue that, in the implicit function theorem, if
$f \in C^k(X \times Y, \R^n)$ for $k \geq 1$, then $g \in C^k(B,A)$.
Conclude that in Problem~5, the dependence is not only $C^1$, but
smooth (i.e., $C^\infty$).

\emph{Hint:} Refer to Cramer's rule for the inverse matrix.

\end{enumerate}
\end{document}




