\documentclass[12pt]{article}

\usepackage[T1]{fontenc}
\usepackage[latin1]{inputenc}
% \usepackage[french]{babel}
\usepackage{amsmath}
\usepackage{amsthm}
%\usepackage{mathrsfs}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}{Definition}[section]
\newenvironment{demo}{\noindent {\bf Dem.}}{\qed}
\newenvironment{remarque}{\noindent {\bf Rem.} \small \itshape}{}
\newenvironment{exemple}{\noindent {\bf Example}}{}

\newcommand{\Lu}{L^1(\Rset)}
\newcommand{\tf}[1]{{\cal F}\left(#1\right)}
\newcommand{\ii}{{\mathrm{i}}}
\newcommand{\Cn}{{\cal C}^{n}}
\newcommand{\dd}{\mathrm{d}}
% ;; \newcommand{\Rset}{{\mathbb R}}
\newcommand{\Rset}{R}
\newcommand{\R}{\mathbb R}
\newcommand{\C}{\mathbb R}
\newcommand{\ex}{\mathrm{e}}
\newcommand{\Cinf}{{\cal C}^{\infty}}
\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\dx}{\dd x}
\newcommand{\ds}{\displaystyle}
\newcommand{\vect}[1]{\overrightarrow{#1}}
\newcommand{\Boule}[2]{\mathscr B(#1,#2)}
\newcommand{\Cercle}[2]{\mathscr C(#1,#2)}
\DeclareMathOperator{\Arg}{Arg}

\newcommand{\dep}[2]{\ds \frac{\partial #1}{\partial #2}}

\title{Example of the \textsf{mdpgd} fonts.}

\author{Paul Pichaureau}


\usepackage[cal=scr,mdpgd,greekfamily = didot]{mathdesign}
%% \usepackage{amssymb}

\begin{document}

\maketitle

\begin{abstract}
    The package \textsf{mdpgd} consists of a full set of
    mathematical fonts, designed to be combined with Adobe
    Adobe Garamond Pro as the main text font.

    This example is extracted from the excellent book {\em
        Mathematics for Physics and Physicists}, {\sc W. Appel},
    Princeton University Press, {\sc 2007}.

\end{abstract}


\section{Conformal maps}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\subsection{Preliminaries}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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Consider a change of variable  $(x,y)\mapsto
(u,v)=\big(u(x,y),v(x,y)\big)$ in the plane $\R^2$, identified
with~$\C$. This change of variable really only deserves the name if
$f$ is locally bijective (i.e., one-to-one); this is the case if the
jacobian of the map is nonzero (then so is the jacobian of the
inverse map):
\begin{equation*}
  \left| \frac{{D}(u,v)}{{D}(x,y)}\right| =
  \begin{vmatrix} 
    \ds\frac{\displaystyle\partial u}{\displaystyle\partial x} &    
    \ds\frac{\displaystyle\partial u}{\displaystyle\partial y} \\[4mm]
    \ds\frac{\displaystyle\partial v}{\displaystyle\partial x} &
    \ds\frac{\displaystyle\partial v}{\displaystyle\partial y} 
  \end{vmatrix}\neq 0
  \qquad\text{and}\qquad
  \left| \frac{{D}(x,y)}{{D}(u,v)}\right|
  =\begin{vmatrix}\ds\dep{x}{u} &\ds \dep{x}{v}\\[4mm]
    \ds\dep{y}{u} &\ds \dep{y}{v} 
  \end{vmatrix}\neq 0.
\end{equation*}
\begin{theorem}
In a complex change of variable
\begin{equation*}
        z= x+\ii y\longmapsto w=f(z)=u+\ii v,
\end{equation*}
and \emph{if $f$ is holomorphic}, then the jacobian of the map is equal to
\begin{equation*}
  J_f(z)=\left| \frac{{D}(u,v)}{{D}(x,y)}\right|=
  \abs{f'(z)}^2.
\end{equation*}
\end{theorem}
\begin{demo}
  Indeed, we have $f'(z)=\dep{u}{x}+\ii\dep{v}{x}$ and hence, by the
  Cauchy-Riemann relations,
  \begin{align*}
    \abs{f'(z)}^2 & = 
    \left(\dep{u}{x}\right)^2+\left(\dep{v}{x}\right)^2 
    =
    \dep{u}{x}\dep{v}{y}-\dep{v}{x}\dep{u}{y}=J_f(z).
  \end{align*}
\end{demo}

\begin{definition}
  \index{Conformal map}%
  \index{Transformation!conformal ---}%
  A \emph{conformal map} or \emph{conformal transformation} of an
  open subset $\Omega\subset\R^2$ into another open subset
  $\Omega'\subset\R^2$ is any map $f:\Omega\mapsto \Omega'$, locally
  bijective, that preserves angles and orientation.
\end{definition}

\begin{theorem}
  Any conformal map is given by a holomorphic function $f$ such
  that the derivative of $f$ does not vanish.
\end{theorem}

This justifies the next definition:
%% ----------------------------------------------------------------------  
\begin{definition}
  \index{Conformal map}%
  \index{Transformation!conformal ---}%
  A \emph{conformal transformation} or \emph{conformal map} of
  an open subset 
  $\Omega\subset\C$ into another open subset
  $\Omega'\subset\C$ is any holomorphic function
  $f:\Omega\mapsto \Omega'$ such that
  $f'(z)\neq 0$ for all $z\in\Omega$.
\end{definition}
%% ----------------------------------------------------------------------  

%% ----------------------------------------------------------------------  
\begin{demo}[that the definitions are equivalent]
  We will denote in general $w=f(z)$.  Consider, in the complex plane, two
  line segments $\gamma_1$ and $\gamma_2$ contained inside the set $\Omega$
  where $f$ is defined, and intersecting at a point $z_0$ in $\Omega$.
  Denote by $\gamma'_1$ and $\gamma_2'$ their images by~$f$.
  
  We want to show that if the angle between $\gamma_1$ and $\gamma_2$ is
  equal to $\theta$, then the same holds for their images, which means that
  the angle between the tangent lines to $\gamma'_1$ and $\gamma'_2$ at
  $w_0=f(z_0)$ is also equal to $\theta$.
  
  Consider a point $z\in\gamma_1$ close to~$z_0$. Its image $w=f(z)$
  satisfies
  \begin{equation*}
    \lim_{z\to z_0} \frac{w-w_0}{z-z_0}=f'(z_0),
  \end{equation*}
  and hence
  $$\displaystyle \lim_{z\to z_0} \Arg
    (w-w_0)-\Arg(z-z_0) = \Arg f'(z_0), $$%
  which shows that the angle between the curve $\gamma'_1$ and the real
  axis is equal to the angle between the original segment $\gamma_1$ and
  the real axis, plus the angle $\alpha=\Arg f'(z_0)$ (which is well
  defined because $f'(z)\neq 0$).
  
  Similarly, the angle between the image curve $\gamma'_2$ and the real
  axis is equal to that between the segment $\gamma_2$ and the real axis,
  plus the same~$\alpha$.
  
  Therefore, the angle between the two image curves is the same as that
  between the two line segments, namely, $\theta$.
  
  Another way to see this is as follows: the tangent vectors of the curves
  are transformed according to the rule $\vect{V}'=\dd f_{z_0}\vect{V}$. But the
  differential of $f$ (when $f$ is seen as a map from $\R^2$ to~$\R^2$) is
  of the form
  \begin{equation}
    \displaystyle \dd f_{z_0}=\begin{pmatrix}
      \displaystyle \dep{P}{x} & \displaystyle \dep{P}{y} \\[4mm]
      \displaystyle \dep{Q}{x} & \displaystyle \dep{Q}{y}\end{pmatrix}
    =
    \abs{f'(z_0)}\begin{pmatrix}\cos\alpha& -\sin\alpha
      \\ \sin\alpha &\cos\alpha \end{pmatrix},
    \label{eq:FSimil}
  \end{equation}
  where $\alpha$ is the argument of $f'(z_0)$. This is the matrix of a
  rotation composed with a homothety, that is, a similitude.
  
  \medskip
%% ······································································  
  % {\begin{picture}(300,100)
      % \put(0,0){\epsfig{file=\Figures/TC.\Ext,height=3.2cm}}
      % \put(20,65){$\gamma_2$} \put(80,55){$\theta$}
      % \put(100,80){$\gamma_1$} \put(195,85){$\gamma'_1$}
      % \put(245,35){$\theta$} \put(270,60){$\gamma'_2$}
    % \end{picture}}
%% ······································································  
  
  Conversely, if $f$ is a map which is $\R^2$-differentiable and preserves
  angles, then at any point $\dd f$ is an endomorphism of~$\R^2$ which
  preserves angles. Since $f$ also preserves orientation, its determinant
  is positive, so $\dd f$ is a similitude, and its matrix is exactly
  as in equation~\eqref{eq:FSimil}. The Cauchy-Riemann equations are
  immediate consequences.
\end{demo}
%% ----------------------------------------------------------------------  

%% ----------------------------------------------------------------------  
\begin{remarque}
  \index{Antiholomorphic function}%
  \index{Function!antiholomorphic ---}%
  An \emph{antiholomorphic} map also preserves angles, but it
  reverses the orientation.
\end{remarque}
%% ----------------------------------------------------------------------  

\newpage
\subsection*{Calcul différentiel}



Pour obtenir la différentielle totale de cette expression, considérée comme fonction de $x$, $y$, ..., donnons à $x$, $y$, ... des accroissements $d\!x$, $d\!y$, .... Soient $\Delta u$, $\Delta v$, ..., $\Delta f$ les accroissements correspondants de $u$, $v$, ...,$f$. On aura
\begin{equation*}
  \Delta f= \dfrac{\partial\! f}{\partial u} \Delta u +  \dfrac{\partial\! f}{\partial v} \Delta v + \hdots + R\Delta u + R_1 \Delta v + \hdots,    
\end{equation*}
$R$, $R_1$, ... tendant vers zéro avec $\Delta u$, $\Delta v$, ....

Mais on a, d'autre part,
\begin{align*}
  \Delta u & =  \dfrac{\partial u}{\partial x} d\! x +   +  \dfrac{\partial u}{\partial y} \Delta y + \hdots + S\Delta x + S_1 \Delta y + \hdots \\
& = du + Sd\! x + S_1 d\! y + \hdots \\
 \Delta v & =  \dfrac{\partial v}{\partial x} d\! x +   +  \dfrac{\partial v}{\partial y} \Delta y + \hdots + T\Delta x + T_1 \Delta y + \hdots \\
& = dv + Td\! x + T_1 d\! y + \hdots \\
\hdots
\end{align*}
$S$, $S_1$, ..., $T$, $T_1$,... tendant vers zéro avec $d\! x$, $d\! y$, ....

Substituant ces valeurs dans l'expression de $\Delta f$, il vient
\begin{equation*}
\begin{array}{rcl}
 \vbox to 25pt {} \Delta f & = &\dfrac{\partial\! f}{\partial u} d u +  \dfrac{\partial\! f}{\partial v} d v + \hdots + \rho d\! x  + \rho_1 d\! y  + \hdots \\
\vbox to 25pt {}& = & \phantom{+} \left( \dfrac{\partial\! f}{\partial u} \dfrac{\partial u}{\partial x} + \dfrac{\partial\! f}{\partial v} \dfrac{\partial v}{\partial x} + \hdots \right) d\! x \\
\vbox to 25pt {}& & + \left( \dfrac{\partial\! f}{\partial u} \dfrac{\partial u}{\partial y} + \dfrac{\partial\! f}{\partial v} \dfrac{\partial v}{\partial y} + \hdots \right) d\! y \\
\vbox to 25pt {}&& + \hdots + \rho d\! x  + \rho_1 d\! y  + \hdots
\end{array}
\end{equation*}
$\rho$, $\rho_1$, ... tendant vers zéro avec $d\! x$, $d\! y$, ....

On aura donc
\begin{align*}
  \dfrac{\partial\! f}{\partial x}& = \dfrac{\partial\! f}{\partial u} \dfrac{\partial u}{\partial x} + \dfrac{\partial\! f}{\partial v} \dfrac{\partial v}{\partial x} + \hdots, \\
  \dfrac{\partial\! f}{\partial y}& = \dfrac{\partial\! f}{\partial u} \dfrac{\partial u}{\partial y} + \dfrac{\partial\! f}{\partial v} \dfrac{\partial v}{\partial y} + \hdots, \\
\hdots
\end{align*}
et, d'autre part,
\begin{equation*}
    df  = \dfrac{\partial\! f}{\partial u} {\mathrm d} u +  \dfrac{\partial\! f}{\partial v}  {\mathrm d} v + \hdots ;
\end{equation*}
d'où les deux propositions suivantes :

{\em La dérivée, par rapport à une variable indépendante $x$, d'une fonction composée $f(u,v,\hdots)$ s'obtient en ajoutant ensemble les dérivées partielles $\dfrac{\partial\! f}{\partial u}$, $\dfrac{\partial\! f}{\partial v}$, ..., respectivement multipliées par les dérivées de $u$, $v$, ... par rapport à $x$.

La différentielle totale $df$ s'exprimer au moyen de $u$, $v$, ..., $du$, $dv$, ..., de la même manière que si $u$, $v$, ... étaient des variables indépendantes.
} 

\hbox to \textwidth { \hfill 
    {\sc Camille Jordan}, {\em Cours d'analyse de l'\'Ecole polytechnique}
}

\end{document}
