%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% M00DT2BC.tex
% Version 1.05 du 3 juin 2000.
% Correction du probl\`eme IIB de la banque PT 2000
%
% R\'edacteur : Claude Mendes <csmendes@club-internet.fr>
% TeXnicien : Daniel Duparc <dduparc@club-internet.fr>
% Lyc\'ee Voltaire
% 101 av. de la R\'epublique
% 75543 PARIS CEDEX 11
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% Merci \`a Yves Saladin pour ses nombreuses am\'eliorations. 
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\title{\textbf{Banque PT 2000 : Math\'ematiques II-B }}

\begin{document}
\author{}
\date{}
\maketitle
%\pagestyle{empty}
%\pagestyle{headings}
%\pagestyle{myheadings}
%\markright{{\bf DUPARC 843}\ \ \ \ \ Correction de Claude {\sc Mendes}}
%\markright{{\bf Banque PT 2000 : Math\'ematiques II-B}}
%\thispagestyle{empty}

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\LARGE \textbf{ Premi\`ere partie}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question I.1
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section*{1. (a) Ensemble des solutions de $ (E) $ }

Par th\'eor\`eme, le syst\`eme diff\'erentiel $ (E) $ poss\`ede une solution unique
$ \overrightarrow{F} $ telle que $ \overrightarrow{F}  (t_{0})  =  \overrightarrow{F_{0}} $
o\`u $ t_{0} $ et $ \overrightarrow{F_{0}} $ sont fix\'es.

%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{1. (b) S'il existe un r\'eel $ t_{0} $ tel que 
$ \overrightarrow{F} (t_{0})=\overrightarrow{0} $ , la
fonction $ \overrightarrow{F} $ est nulle}

D\'esignons par $ \oa{\Theta}  $ 
la fonction nulle de \R{} vers $ \R ^{3} $ .
Cette fonction est solution de $ (E) $ et 
$ \oa{\Theta} (t_{0})=\overrightarrow{0} $.
Donc si $ \overrightarrow{F} $ est solution de $ (E) $ telle que 
$ \overrightarrow{F} (t_{0})=\overrightarrow{0} $, 
d'apr\`es la question 1.a, $ \oa{\Theta } $
et $ \overrightarrow{F} $ sont confondues.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question I.2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\section*{2. (a) $ \nu  $ et $ \overrightarrow{f} $
sont des fonctions d\'erivables }

D\'esignons par $ f_{1} $ , $ f_{2} $ , $ f_{3} $ les coordonn\'ees de
$ \overrightarrow{F} $ dans la base $ B_{0} $ . Comme $ \overrightarrow{F}$
est de classe $ {\mathcal{C}}^{1} $ sur $ \R  $, les trois fonctions
coordonn\'ees $ f_{1} $ , $ f_{2} $ , $ f_{3} $ sont aussi de classe
$ {\mathcal{C}}^{1} $ et par suite $ \nu =\sqrt{f_{1}^{2}+f_{2}^{2}+f_{3}^{2}} $
aussi.

Comme $ \overrightarrow{F} $ ne s'annule pas sur $ \R  $ , $ \nu  $
ne s'annule pas sur $ \R  $ et donc  $ \frac{1}{\nu } $ sera de classe
$ {\mathcal{C}}^{1} $ sur $ \R  $ . $ \overrightarrow{f} $ \'etant le
produit de la fonction vectorielle $ \overrightarrow{F} $ et de $ \frac{1}{\nu } $,
toutes deux de classe $ {\mathcal{C}}^{1} $ , est une fonction vectorielle
de classe $ {\mathcal{C}}^{1} $ sur $ \R  $ .

%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{2. (b) $ \overrightarrow{f}(t) $ et $ \overrightarrow{f'(t)} $
sont deux vecteurs orthogonaux pour tout r\'eel $ t $ }

$ ||  \overrightarrow{f} (t)  ||  =1 $ et donc 
$ (  \overrightarrow{f}  )^{2}(t)=1 $
. 

Si nous d\'erivons, nous obtenons 
$$ 2   \overrightarrow{f} (t) 
 . \overrightarrow{f'}(t) =0 $$
donc
$ \overrightarrow{f}(t) $ et $ \overrightarrow{f'(t)} $ sont deux vecteurs
orthogonaux pour tout r\'eel $ t $.

%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{2. (c) Calcul de la d\'eriv\'ee de $ \nu  $ }

$ \nu   (t)  =  \sqrt{(\overrightarrow{F}(t))^{2}} $ , donc

$$ \nu'(t) = {\overrightarrow{F(t)} \cdot \overrightarrow{F'(t)}\over \nu (t)} = 
\overrightarrow{f(t)} \cdot \overrightarrow{F'(t)}
$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question I.3
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{3. La fonction vectorielle satisfait \`a }
$${d\overrightarrow{f}\over dt}
 =  a  (\overrightarrow{f})  -  ( a(\overrightarrow{f}) \cdot 
\overrightarrow{f} ) \oa{f}$$
Nous avons :
$$\oa{F'(t)} = a(\oa{F}) = a(\nu (t) . \oa{f(t)}) = \nu (t) . a(\oa{f(t)})$$
\`a cause de la lin\'earit\'e de $a$.
D'autre part :
$$\oa{F(t)} = \nu (t) . \oa{f(t)}
$$
d'o\`u 
$$ \oa{F'(t)}= \nu' (t) . \oa{f(t)} +\nu (t) . \oa{f(t)}
$$
et donc :
$$ \nu (t) . a(\oa{f(t)})= \Bigl( \oa{f(t)} \cdot \oa{F' (t)}\Bigr) 
. \oa{f (t)}+ \nu (t) . \oa{f' (t)}
$$
en vertu de la question 2. (b). Rempla\c{c}ons $\oa{F' (t)}$ :
$$ \nu (t) . a(\oa{f(t)})= \Bigl( \oa{f(t)} \cdot a\Bigl( \nu (t). \oa{f
(t)}\Bigr)\Bigr) 
. \oa{f (t)}+ \nu (t) . \oa{f' (t)}
$$
En utilisant la lin\'earit\'e de $a$ et la bilin\'earit\'e du 
produit scalaire, nous en d\'eduisons :
$$ \nu (t) . a(\oa{f(t)})= \nu (t) \Bigl( \oa{f(t)} \cdot a\Bigl( \oa{f
(t)}\Bigr)\Bigr) 
. \oa{f (t)}+ \nu (t) . \oa{f' (t)}
$$
Nous pouvons simplifier par $\nu (t)$ qui est toujours 
non nul :
$$\overrightarrow{f' (t)}
 =  a  (\overrightarrow{f(t)})  -  ( a(\overrightarrow{f(t)}) \cdot 
\overrightarrow{f(t)} ) \oa{f(t)}$$
donc :
$${d\overrightarrow{f}\over dt}
 =  a  (\overrightarrow{f})  -  ( a(\overrightarrow{f}) \cdot 
\overrightarrow{f} ) \oa{f}$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\LARGE \textbf{ Deuxi\`eme partie}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question II.1
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section*{1. (a) Expression de $z(t)$ }
Le syst\`eme $(E)$ peut \^etre \'ecrit :

$$
\begin{array}{llll}
x'(t)&=&                    & G(\lambda -1) \ y(t)\\
y'(t)&=& G(\lambda +1) \ x(t) &                   \\
z'(t)&=& 0                  &                  
\end{array}
$$
Donc $z$ est constant :
$$ z(t) = z_0
$$ 

%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{1. (b) Int\'egration de $(E)$ lorsque $x_0 = y_0 = 0$}  
La fonction $\oa{F}$ telle que $\oa{F(t)} = z_0 \oa{e_3}$ 
est une solution de $(E)$ v\'erifiant les conditions initiales.
Comme cette solution est unique, c'est la solution cherch\'ee.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question II.2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section*{2. D\'emonstration de $r^2 x(t) x'(t) + y(t) y'(t) =0$}
On vient de voir que :
$x'(t)=                     G(\lambda -1) \ y(t)$ et
$y'(t)= G(\lambda +1) \ x(t)$.
Donc :
$$ r^2 x(t) x'(t) + y(t) y'(t) = 
\Bigl[ r^2 G(\lambda -1)+G(\lambda+1)\Bigr] x(t) y(t)
$$
$$ r^2 x(t) x'(t) + y(t) y'(t) = 
 G\Bigl[{1+\lambda \over 1-\lambda}(\lambda -1)+(\lambda +1)) \Bigr] 
x(t) y(t) =0
$$

%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Courbes int\'egrales}
On tire de ce qui pr\'ec\`ede que $r^2 x^2(t) + y^2(t) $ 
est constant :
$$r^2 x^2(t) + y^2(t) = r^2 x^2_0 + y^2_0$$
On obtient donc en int\'egrant les  \'equations :
$$
\left\{
\begin{array}{lll}
z & = & z_0\\
{x^2 \over ({\mu \over r})^2}+{y^2 \over \mu^2} & =& 1
\end{array}
\right.
$$
avec $\mu = \sqrt{r^2 x^2_0 + y^2_0}$.

Les courbes int\'egrales sont donc incluses dans 
des ellipses situ\'ees 
dans des plans  "horizontaux", centr\'ees sur 
l'axe des $z$, dont les axes sont des 
parall\`eles aux axes de coordonn\'ees 
et dont le rapport des axes est fixe. Pour $z_0$ fix\'e,
ces ellipses sont homoth\'etiques les unes des autres.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question II.3
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section*{3. (a) $x$ et $y$ sont solutions de l'\'equation
diff\'erentielle $$u'' + \omega^2 u =0$$}

D'apr\`es les deux premi\`eres \'equations du syst\`eme $(E)$,
$$x'(t) = G(\lambda -1)\ y(t)
$$
$x'$ est de classe ${\mathcal C}^1$ puisque $y$ 
est de classe  ${\mathcal C}^1$ et 
donc :
$$ x''(t) = G(\lambda -1) \ y'(t) = G^2(\lambda^2 - 1) x(t)
$$
et :
$$y'(t) = G(\lambda +1)\ x(t)
$$
donc :
$$ y''(t) = G(\lambda +1) \ x'(t) = G^2(\lambda^2 - 1) y(t)
$$

%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{3. (b) Int\'egration de $(E)$ lorsque $x^2_0 + y^2_0 \neq 0$}

$x$ est de la forme :
$$ x(t) = \alpha \cos (\omega t) +\beta \sin (\omega t)
$$
o\`u $\alpha$ et $\beta$ sont des constantes r\'eelles.

$x(0) = \alpha =  x_0$, donc 
$x(t) = x_0 \cos (\omega t) +\beta \sin (\omega t)
$.

$x'(t) = -\omega x_0 \sin (\omega t) +\beta \omega \cos (\omega t)
= G (\lambda -1) y(t)$

Nous en d\'eduisons  
$$ x'(0) = \omega \beta = G (\lambda -1) y_0$$ 
et 
$$\beta ={G(\lambda -1)\over \omega }y_0 = -{y_0\over r}$$
Donc :
$$x(t) = x_0 \cos (\omega t) -{y_0\over r} \sin (\omega t)
$$
$$ y(t) = {1\over G (\lambda -1)} x' (t)
$$
Nous en d\'eduisons  :
$$y(t) = y_0 \cos (\omega t) +{r x_0} \sin (\omega t)
$$

%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{3. (c)  Si $x_0 \neq 0$, il existe une constante $t_0$ 
telle que~:
$$ {y(t)\over x(t)}= r \tan \omega (t - t_0)
$$}
D'apr\`es la question pr\'ec\'edente, 
$${y(t)\over x(t)}={y_0 \cos (\omega t) +{r x_0} \sin (\omega t)
\over x_0 \cos (\omega t) -{y_0\over r} \sin (\omega t)
} =
r{{y_0\over r x_0} +\tan (\omega t)\over
1- {y_0\over r x_0}\tan (\omega t)}
$$
au moins dans un voisinage de $t=0$, car $x_0$ est
suppos\'e non nul (continuit\'e).
Donc il existe un  r\'eel $t_0$ de l'intervalle 
$$\ds \Bigl] -{\pi \over 2 \omega }, {\pi \over 2 \omega }\Bigr[$$
tel que 
$$ \tan (\omega t_0) = -{y_0\over r x_0}
$$
et on a alors :
$$ {y(t)\over x(t)}= r \tan \omega (t - t_0)
$$
Remarque.

Sans condition sur $x_0$, on pouvait \'etablir qu'avec ce m\^eme
$t_0$, on avait pour tout  $t$ de $\R$,
$x(t)=\sqrt{x_0^2 r^2+y_0^2}\cos \omega(t-t_0)$ et $y(t)=r \sqrt{x_0^2
r^2+y_0^2}\sin \omega(t-t_0)$, ce qui montrait  que les ellipses
\'etaient parcourues enti\`erement.

\vskip 5mm
\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\LARGE \textbf{ Troisi\`eme partie}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question III.1
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section*{1. (a) Figure illustrant la d\'efinition de $\theta (t)$ et 
$\phi (t)$}

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

D\'esignons par $\oa{f_1(t) }$, $\oa{f_2(t) }$, $\oa{f_3(t) }$
les composantes respectives de $\oa{f(t) }$ sur les 
droites vectorielles $\hbox{Vect} (\oa{e_1})$,
$\hbox{Vect} (\oa{e_2})$, $\hbox{Vect} (\oa{e_3})$.

$\theta (t)$ est l'angle de $\oa{f(t) }$ avec $\oa{e_3}$,
cet angle \'etant un angle g\'eom\'etrique dont la mesure
va de $0$ \`a $\pi$.

$\phi (t)$ est l'angle de $\oa{f_1(t) }+\oa{f_2(t) }$,
projection orthogonale de $\oa{f(t) }$ sur le plan 
$\hbox{Vect} ( \oa{e_1}, \oa{e_2})$,
avec $\oa{e_1}$. Cet angle est orient\'e et appartient 
\`a un intervalle de longueur $2\pi$. 

En fait, il y a une infinit\'e de r\'eponses \`a cette
question, et on ne comprendra que dans la question
5. (c) de la troisi\`eme partie quelle est la r\'eponse 
 "attendue" par le concepteur du probl\`eme !

%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{1. (b) S'il existe $t_0$ tel que $\sin \theta (t_0) =0$
alors $\sin \theta $ est la fonction nulle}

S'il existe  $t_0$ tel que $\sin \theta (t_0) =0$,
alors $\oa{F(t_0)} = z_0 \oa{e_3}$ et, d'apr\`es la question 1. (b)
de la deuxi\`eme partie, pour tout $t$,
$$\oa{F(t)} = z_0 \oa{e_3}$$
et donc $\oa{f(t)}$ est  $\oa{e_3}$ si $z_0 >0$ et est $-\oa{e_3}$
si $z_0 <0$. 

$\theta (t)$ est constant et vaut soit $0$, soit $\pi$, donc la fonction
$t \mapsto \sin (\theta (t))$ est identiquement nulle.

%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{1. (c) S'il existe $t_0$ tel que $\cos \theta (t_0) =0$
alors $\cos \theta $ est la fonction nulle}

D'apr\`es la question 1. (a) de la deuxi\`eme partie, $z(t) = z_0$.
Donc si $z_0$ est nul, pour tout $t$, 
$t \mapsto \cos (\theta (t))$ est identiquement nulle.

\section*{1. (d) S'il existe $t_0$ tel que $\sin 2 \theta (t_0) =0$
alors $ \theta $ est une fonction constante}

S'il existe $t_0$ tel que $\sin 2 \theta (t_0) =0$, alors
$2 \sin \theta (t_0) \cos \theta (t_0) =0$, donc on a 
$ \sin \theta (t_0) =0$
ou bien $ \cos \theta (t_0) =0$.

On se trouve donc dans un (et un seul) des deux cas pr\'ec\'edents,
donc $\theta$ est une fonction constante.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question III.2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{2. (a) \'Ecriture de $\oa{f'(t)}$ en fonction de $\theta (t)$,
$\phi (t)$, $\theta' (t)$ et $\phi' (t)$}

$$ \oa{f(t)} = \sin \bigl( \theta (t)\bigr)\  \oa{u(t)} + 
\cos \bigl( \theta (t)\bigr)\   \oa{e_3} 
$$
$$ \oa{u'(t)} = -\sin \bigl( \phi (t)\bigr)\  \phi'(t)\  \oa{e_1}
+ \cos \bigl( \phi (t)\bigr)\  \phi'(t)\  \oa{e_2}
= \phi' (t)\  \oa{v(t)}
$$
Nous en d\'eduisons :
$$\oa{f'(t)} = \cos \bigl( \theta (t)\bigr)\  \theta'(t)\ \oa{u(t)} +
\sin \bigl( \theta (t)\bigr)\  \phi'(t)\ \oa{v(t)} 
-\sin \bigl( \theta (t)\bigr)\  \theta'(t)\ \oa{e_3} 
$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{2. (b) Matrice $A_\phi (t) $ de $a$ dans la base $B_\phi (t)$}

D\'esignons par $P$ la matrice de passage de la base $B_0$ \`a la base
$B_\phi (t)$.
$$P=
\left(
\begin{array}{ccc}
\cos \phi (t) & -\sin \phi (t) & 0 \\
\sin \phi (t) & \cos \phi (t)  & 0 \\
0             &       0        & 1
\end{array}
\right)
$$
Comme $B_\phi (t)$ est une base orthonormale, $P$
est une matrice orthogonale r\'eelle~:
$$P^{-1}={}^tP=
\left(
\begin{array}{ccc}
\cos \phi (t) & \sin \phi (t)   & 0 \\
-\sin \phi (t)& \cos \phi (t)   & 0 \\
0             &    0            & 1
\end{array}
\right)
$$
$$A_{\phi (t)}= P^{-1}AP=
\left(
\begin{array}{ccc}
\cos \phi (t) & \sin \phi (t)   & 0 \\
-\sin \phi (t)& \cos \phi (t)   & 0 \\
0             &   0             & 1
\end{array}
\right)
\left(
\begin{array}{ccc}
0              & G(\lambda -1)   & 0 \\
G(\lambda +1 ) & 0               & 0 \\
0              & 0               & 0
\end{array}
\right)
P
$$
$$A_{\phi (t)}=
\left(
\begin{array}{ccc}
G (\lambda +1) \sin \phi (t)    & G (\lambda -1) \cos \phi (t)   & 0 \\
G (\lambda +1) \cos \phi (t)    & -G (\lambda -1) \sin \phi (t)  & 0 \\
0                               &   0                            & 0
\end{array}
\right)
\left(
\begin{array}{ccc}
\cos \phi (t) & -\sin \phi (t) & 0 \\
\sin \phi (t) & \cos \phi (t)  & 0 \\
0             &       0        & 1
\end{array}
\right)
$$
$$A_{\phi (t)}=
\left(
\begin{array}{ccc}
2G \lambda \sin \phi (t) \cos \phi (t)           & 2 G \lambda  \cos^2 \phi (t)  -G (1+\lambda ) & 0 \\
2 G \lambda  \cos^2 \phi (t)  +G (1-\lambda )    & -2G \lambda \sin \phi (t) \cos \phi (t)   & 0 \\
0                               &   0                            & 0
\end{array}
\right)
$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{2. (c) Calcul de $a\bigl( \oa{f(t)} \bigr)\cdot\oa{f(t)}$}

La matrice de $a\bigl( \oa{f(t)} \bigr)$ dans la base $B_{\phi} (t)$
est :
$$
{\mathcal M}_{B_{\phi} (t)}\bigl( a(\oa{f(t)})\bigr)= A_{\phi (t)}\times 
{\mathcal M}_{B_{\phi} (t)}\bigl( \oa{f(t)}\bigr) 
$$
$$
{\mathcal M}_{B_{\phi} (t)}\bigl( a(\oa{f(t)})\bigr)= 
\left(
\begin{array}{ccc}
2G \lambda \sin \phi (t) \cos \phi (t)           & 2 G \lambda  \cos^2 \phi (t)  -G (1+\lambda ) & 0 \\
2 G \lambda  \cos^2 \phi (t)  +G (1-\lambda )    & -2G \lambda \sin \phi (t) \cos \phi (t)   & 0 \\
0                               &   0                            & 0
\end{array}
\right)
\left(
\begin{array}{c}
\sin \theta (t)  \\
0                \\
\cos \theta (t)  
\end{array}
\right)
$$ 
$$
{\mathcal M}_{B_{\phi} (t)}\bigl( a(\oa{f(t)})\bigr)= 
\left(
\begin{array}{c}
2G \lambda   \sin \phi (t)    \cos \phi (t)\sin \theta (t)  \\
2G \lambda   \cos^2 \phi (t)\sin \theta (t)  + (1-\lambda) G \sin \theta (t)\\
0 
\end{array}
\right)
$$
On en d\'eduit, en effectuant le calcul dans la base $B_{\phi} (t)$ :
$$a\bigl( \oa{f(t)} \bigr)\cdot\oa{f(t)} =
2G \lambda   \sin \phi (t)    \cos \phi (t)\sin^2 \theta (t)
$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question III.3
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{3. (a) Syst\`eme diff\'erentiel $(S)$ satisfait par $\theta$ et 
$\phi$ }
Dans la question 3 de la premi\`ere partie, nous avons vu que :
$${d\overrightarrow{f}\over dt}
 =  a  (\overrightarrow{f})  -  ( a(\overrightarrow{f}) \cdot 
\overrightarrow{f} ) \oa{f}$$
En exprimant les coordonn\'ees des vecteurs des deux membres dans 
la base $B_\phi (t)$, nous en d\'eduisons le syst\`eme :
$$\left\{
\begin{array}{ccc}
\cos \theta (t)\  \theta' (t)     &=& 2G \ \lambda\ \sin\phi (t)\ \cos\phi
(t)\ \sin \theta (t)
-2G\  \lambda\ \sin\phi (t)\ \cos\phi (t)\ \sin^3 \ \theta (t)\\
\sin \theta (t)\  \phi' (t)     &=& 2G \ \lambda\ \cos^2\phi (t)\ \sin \ \theta (t)+
\ (1-\lambda )\ G \ \sin \theta (t)\\
-\sin \theta (t)\  \theta' (t)     &=& -2G\  \lambda\ \sin\phi (t)\ \cos\phi
(t)\  \sin^2 \theta (t)\ 
\cos \theta (t)
\end{array}
\right.
$$
que l'on peut \'ecrire encore :
$$\left\{
\begin{array}{ccc}
\cos \theta (t)\  \theta' (t)     &=& 2G \ \lambda\ \sin\phi (t)\ \cos\phi (t)
\ \sin \theta (t)\ \cos^2 \theta (t)\\
\sin \theta (t)\  \theta' (t)     &=& 2G\  \lambda\ \sin\phi (t)\ \cos\phi (t)
\ \sin^2 \theta (t)\ \cos \theta (t)\\
\sin \theta (t)\  \phi' (t)     &=&
[ 2G\  \lambda\ \cos^2\phi (t)+
(1-\lambda )\ G ] \ \sin \theta (t) 
\end{array}
\right.
$$
Les deux premi\`eres \'equations sont \'equivalentes \`a :
$$\theta' (t)   = 2G \ \lambda\ \sin\phi (t)\ \cos\phi (t)
\ \sin \theta (t)\ \cos \theta (t)
$$
donc le syst\`eme est \'equivalent \`a :
$$\left\{
\begin{array}{ccc}
\theta' (t) - 2G \ \lambda\ \sin\phi (t)\ \cos\phi (t)
\ \sin \theta (t)\ \cos \theta (t)&=&0\\
\sin \theta (t)\  \bigl(\phi' (t) - [ 2G\  \lambda\ \cos^2\phi (t)+
(1-\lambda )\ G ] \bigr)   &=& 0
\end{array}
\right.
$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{3. (b) Si $\oa{f_0}$ et $\oa{e_3}$ ne sont pas
colin\'eaires, le syst\`eme $(S)$ est \'equivalent \`a :}
\begin{eqnarray}
\phi' &=& 2 G \lambda \cos^2 \phi + G (1- \lambda )\\
\theta'&=& 2 G \lambda \sin \phi \cos \phi \sin \theta \cos \theta 
\end{eqnarray}

Si $\oa{f_0}$ et $\oa{e_3}$ ne sont pas colin\'eaires,
d'apr\`es la question 1. (b) de la troisi\`eme partie, 
$\oa{f (t)}$ et $\oa{e_3}$ ne le sont jamais,  $\sin \theta (t) \neq 0$,
et donc le syst\`eme $(S)$ est \'equivalent \`a : 
\setcounter{equation}{0}
\begin{eqnarray}
\phi' &=& 2 G \lambda \cos^2 \phi + G (1- \lambda )\\
\theta'&=& 2 G \lambda \sin \phi \cos \phi \sin \theta \cos \theta 
\end{eqnarray}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question III.4
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{4. Int\'egration de $(S)$ lorsque $\lambda = 0$ et trajectoire du
point $m$}
Lorsque $\lambda = 0$, le syst\`eme $(S)$ devient :
\setcounter{equation}{2}
\begin{eqnarray}
\phi' &=& G\\
\theta'&=& 0
\end{eqnarray}
et donc :
$$\left\{
\begin{array}{ccc}
\phi (t)&=& G (t-t_0)\\
\theta (t)&=& \theta_0
\end{array}
\right.
$$
La trajectoire du point $m$ tel que $\oa{Om (t)} = \oa{f(t)}$
est le cercle de la sph\`ere unit\'e de centre $O$, dont la distance
z\'enithale est $\theta_0$. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question III.5
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{5. (a) \'Etude des solutions $\phi $ de (1)
quand $\lambda $ est non nul}
$$2 G \lambda \cos^2 \phi + G (1- \lambda )
=G[2\lambda \cos^2 \phi +1-\lambda]
= G[(2\cos^2 \phi -1)\lambda +1]
$$
$$2 G \lambda \cos^2 \phi + G (1- \lambda )=
G \ [\lambda \cos (2\phi ) +1]
$$
Donc toute solution $\phi$ de (1) est telle que :
$$(1+|\lambda | )\  G  \ge \phi' (t) \ge (1-|\lambda | ) \ G >0
$$
car $-1 <\lambda <1$.

Donc $\phi$ est strictement croissante.

Donc pour tout $t>0$,
$$\phi (t) - \phi (0) =
\int^t_0 \phi'(s) ds \ge
\int^t_0 (1-|\lambda | ) \ G \ ds = (1- |\lambda |) \ G t
$$
On en d\'eduit que $\ds \lim_{+\infty} \phi = +\infty $.

De m\^eme  pour tout $t<0$,
$$\phi (t) - \phi (0) =
-\int_t^0 \phi'(s) \ ds \le
-\int_t^0 (1- |\lambda |) \ G \ ds = (1-|\lambda |) \ G t
$$
On en d\'eduit que $\ds \lim_{-\infty} \phi = -\infty $.

En r\'esum\'e, sur $]-\infty ,+ \infty[$, $\phi$ 
est une fonction continue et strictement
croissante telle que $\ds \lim_{+\infty} \phi = +\infty $ et
$\ds \lim_{-\infty} \phi = -\infty $.

Donc $\phi$ r\'ealise une bijection de \R{} dans \R .

%%%%%%%%%%%%%%%%%%%%%%
\section*{5. (b) $\theta$ reste constant le long d'une 
courbe int\'egrale si et seulement si $\sin 2\theta_0 =0$}

D'apr\`es la question pr\'ec\'edente, $\sin \phi \cos \phi$ 
n'est pas la fonction nulle. Pour que $\theta' (t)$ 
soit toujours nul, il faut et il suffit que 
$\sin \theta \cos \theta $ soit constamment nul, c'est-\`a-dire 
que $\sin 2\theta$ le soit. D'apr\`es les questions
1. (b), 1. (c) et 1. (d) de la troisi\`eme partie,
pour qu'il en soit ainsi, il suffit que ce soit le 
cas pour une valeur $t_0$.

%%%%%%%%%%%%%%%%%%%%%%
\section*{5. (c) Int\'egration de (2) lorsque $\sin 2\theta_0 \neq 0$, 
\`a l'aide de (1)}

De (1) et (2) nous tirons :
$$ {d\phi\over d\theta} =
{2\lambda \cos^2\phi + 1-\lambda
\over 2\lambda \sin \phi  \cos \phi \sin \theta \cos \theta}
$$
soit :
$${2\ \lambda \ \sin \phi\  \cos \phi \ d\phi \over 
2\  \lambda \ \cos^2 \phi + 1-\lambda} =
{d\theta \over \sin \theta \cos \theta
}$$
qui est \`a variables s\'epar\'ees :
$$-{1\over 2}\ln ( 2\  \lambda \ \cos^2 \phi + 1-\lambda )
= \ln |\tan \theta | + C_1
$$ 
qui \'equivaut \`a :
$$ \tan \theta = 
{C_2
\over \sqrt{2\  \lambda \ \cos^2 \phi + 1-\lambda}}
$$
Or :
$$2\  \lambda \ \cos^2 \phi + 1-\lambda
= 2\  \lambda \ \cos^2 \phi + ( 1-\lambda ) (\cos^2 \phi + \sin^2 \phi )$$
$$2\  \lambda \ \cos^2 \phi + 1-\lambda
=  (\lambda +1 ) \cos^2 \phi + (1- \lambda ) \sin^2 \phi
= (1-\lambda ) [\sin^2 \phi + r^2 \cos^2 \phi ]
$$
On obtient donc :
$$ \tan \theta = {C\over
\sqrt{\sin^2 \phi + r^2 \cos^2 \phi  }}
$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question III.6
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{6. (a) $\phi_2 : t\mapsto \phi_1 (t-t_1) $ est 
aussi une solution de (1)}

Comme $\phi_2 (t) = \phi_1 (t-t_1)$, 
$$\phi'_2 (t) = \phi'_1 (t-t_1)
= 2 G \lambda \cos^2 \phi_1 (t-t1) + G (1-\lambda)=
2 G \lambda \cos^2 \phi_2 (t) + G (1-\lambda )$$
Donc $\phi_2$ est solution de (1).

%%%%%%%%%%%%%%%%%%%%%%
\section*{Propri\'et\'e g\'eom\'etrique de l'ensemble ${\mathcal C}_\lambda$}
Soit $M_2$ le point de ${\mathcal C}_2$ ayant pour abscisse $t$.
Il a pour ordonn\'ee $\phi_2(t) = \phi_1(t-t_1)$. Il est l'image par
la translation de vecteur $t_1 \oa{i}$ du point $M_1$ de 
${\mathcal C}_1$ d'abscisse $t-t_1$, $(\oa{i},\oa{j})$ \'etant
une base du plan, les $t$ \'etant en abscisses, les $\phi (t)$
\'etant en ordonn\'ees, la base \'etant suppos\'ee orthonormale.

Par suite, l'image par une translation de vecteur $t_1 \oa{i}$,
o\`u $t_1$ d\'ecrit \R{}, de toute courbe int\'egrale de (1) est 
une autre courbe int\'egrale de (1).

Autrement dit, en revenant aux courbes int\'egrales,
la transform\'ee par une rotation d'axe $Oz$ d'une courbe
int\'egrale de ${\mathcal C}_\lambda$ est aussi une courbe
int\'egrale de ${\mathcal C}_\lambda$.

%%%%%%%%%%%%%%%%%%%%%% 
\section*{6. (b) $\phi_3 : t\mapsto -\phi_1 (-t) $ est 
aussi une solution de (1)}

Comme $\phi_3 (t) = -\phi_1 (-t)$, 
$$\phi'_3 (t) = \phi'_1 (-t)
= 2 G \lambda \cos^2 \phi_1 (-t) + G (1-\lambda)=
2 G \lambda \cos^2 (-\phi_3 (t)) + G (1-\lambda )$$
Donc $\phi_3$ est solution de (1).

%%%%%%%%%%%%%%%%%%%%%%
\section*{Propri\'et\'e g\'eom\'etrique de l'ensemble ${\mathcal C}_\lambda$}
Soit $M_3$ le point de ${\mathcal C}_3$ ayant pour abscisse $t$.
Il a pour ordonn\'ee $\phi_3(t) = -\phi_1(-t)$. Il est l'image par
la compos\'ee des sym\'etries par rapport aux axes,  du point $M_1$ de 
${\mathcal C}_1$ d'abscisse $-t$, 
les $t$ \'etant en abscisses, les $\phi (t)$
\'etant en ordonn\'ees.

Par suite, l'image par la sym\'etrie par rapport \`a l'origine
de toute courbe int\'egrale de (1) est 
une autre courbe int\'egrale de (1).

Autrement dit, en revenant aux courbes int\'egrales,
la transform\'ee par une sym\'etrie orthogonale
par rapport au plan $xOz$  d'une courbe
int\'egrale de ${\mathcal C}_\lambda$ est aussi une courbe
int\'egrale de ${\mathcal C}_\lambda$.

%%%%%%%%%%%%%%%%%%%%%% 
\section*{6. (c) $\phi_4 : t\mapsto {\ds \pi\over \ds 2}-\phi_1 (-t) $ est 
solution de l'\'equation (1) associ\'ee au param\`etre $-\lambda$}

Comme $\phi_4 (t) = {\ds \pi\over \ds 2}-\phi_1 (-t)$, 
$$\phi'_4 (t) = \phi'_1 (-t)
= 2 G \lambda \cos^2 \phi_1 (-t) + G (1-\lambda)$$
$$\phi'_4 (t) 
= 2 G \lambda \cos^2 ({\ds \pi\over \ds 2}-\phi_4 (t)) + G (1-\lambda)$$
$$\phi'_4 (t) 
= 2 G \lambda \sin^2 \phi_4 (t) + G (1-\lambda )$$
$$\phi'_4 (t) 
= 2 G \lambda (1-\cos^2 \phi_4 (t)) + G (1-\lambda )$$
$$\phi'_4 (t) 
= 2 G (-\lambda) \cos^2 \phi_4 (t) + G (1-(-\lambda ))$$
Donc $\phi_4$ est solution de l'\'equation (1) associ\'ee au
param\`etre $-\lambda$.

%%%%%%%%%%%%%%%%%%%%%%
\section*{Relation entre ${\mathcal C}_\lambda$ et
${\mathcal C}_{-\lambda}$}

On passe du point $(-t,\phi_1 (-t))$ au point $(t,\phi_1 (-t))$
par une sym\'etrie par rapport \`a l'axe des ordonn\'ees.

On passe du point $(t,\phi_1 (-t))$ au point 
$(t,{\pi\over 2}-\phi_1 (-t))$ par une sym\'etrie par rapport
\`a la parall\`ele \`a l'axe des abscisses d'ordonn\'ee ${\pi \over 4}$,
donc on passe du point  $(-t,\phi_1 (-t))$ au point
$(t,\phi_4 (t))$
par une sym\'etrie par rapport au point $I (0,{\pi \over 4})$.

Donc ${\mathcal C}_{-\lambda}$ se d\'eduit de ${\mathcal C}_{\lambda}$
par une sym\'etrie de centre $I$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question III.7
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{7. (a) et (b) Int\'egration de (1) sur $]t_k , t_{k+1}[$ 
par le changement de variable $u = \tan \phi$}

$u = \tan \phi$, $du = {\ds d\phi \over \ds \cos^2 \phi}$,
d'o\`u $d\phi = {\ds du\over \ds 1+u^2}$,
$${du\over 1+u^2} =
\Bigl[ {2 G \lambda\over 1+u^2} + G (1-\lambda )\Bigr] dt
$$
donc :
$$dt = {du \over 2G \lambda + G (1-\lambda )(1+u^2)}
$$
$$dt = {du \over G (1-\lambda) u^2 + G (1+\lambda ) }
$$
$$dt = {1\over G (1-\lambda)} \ {du \over u^2+r^2}
$$
$$dt = {1\over G r (1-\lambda)} \ {d({u\over r}) \over ({u\over r})^2+1}
$$
$$ t= {1\over G r (1-\lambda)} \hbox{\textrm Arctan} ({u\over r})
+ \tau_k
$$
$\tau_k$ est la valeur de $t$ obtenue lorsque $u=0$, 
c'est-\`a-dire lorsque $\tan \phi = 0$ et 
$$\phi \in \Bigl] {\pi\over 2} + k \pi\  ,\  {\pi\over 2} + (k+1) \pi\Bigr[
$$
donc lorsque $\phi = (k+1) \pi$.

%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{7. (c) $\oa{f}$ est une fonction p\'eriodique }

D'apr\`es la question 5. (a) de la troisi\`eme partie,
$\phi$ est de classe ${\mathcal C}^1$ sur $\R $. La fonction
$\phi \mapsto u$ est p\'eriodique de p\'eriode $\pi$ :
$$\phi = (k+1) \pi + \hbox{\textrm Arctan} (u)
$$
$${u\over r}= \tan \bigl( Gr (1-\lambda ) (t-\tau_k )\bigr)
$$
d'o\`u
$$ u = r \tan \bigl( Gr (1-\lambda ) (t-\tau_k )\bigr)
$$
et la fonction $t \mapsto u$ est p\'eriodique de p\'eriode :
$${\pi \over G r (1-\lambda )}
$$ 
ce qui correspond \`a un accroissement de $\pi$ pour $\phi$.

La p\'eriode $T$ de $\oa{f}$ correspond \`a un accroissement
de $2\pi$ de $\phi$, donc :
$$ T= {2\pi \over G r (1-\lambda )}
.$$
Or
$$ {\pi\over G}(r+{1\over r})= {\pi\over G}{r^2+1\over r}
={\pi\over G}{{1+\lambda \over 1-\lambda}+1\over r }
={2\pi\over Gr(1-\lambda)}
$$
et d'autre part :
$${2\pi\over Gr(1-\lambda)} =
{2\pi \over G \sqrt{{1+\lambda\over 1-\lambda}}(1-\lambda )}
={2\pi \over G \sqrt{1-\lambda^2}}
={2\pi \over \omega}
$$
$$T= {\pi\over G}(r+{1\over r})={2\pi \over \omega}
$$


\end{document}
