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\pagenumbering{Alph}
\section{Annexe}

\subsection{Question 2}
\vspace*{-2em}
\begin{align*}
	\left\{\begin{aligned}
		x_1 &\approx \phi^Tx_{1d}(t) \\
		x_2 &\approx \phi^Tx_{2d}(t)
	\end{aligned}\right. & &
	\left\{\begin{aligned}
		e_1 &\approx \phi^Te_{1d}(t) \\
		e_2 &\approx \phi^Te_{2d}(t)
	\end{aligned}\right.
\end{align*}

En utilisant la première ligne de l'équation 1, on trouve :

\begin{equation*}
	\int\phi(\zeta)d\zeta\times\phi^T\dot{x_{1d}} = \int\phi(\zeta)d\zeta\times\frac{\partial^2}{\partial\zeta^2} = \ddot{\phi}(\zeta)^Te_{2d}
\end{equation*}

\begin{equation}
	\underbrace{\int_{0}^{L}\phi(\zeta)\phi(\zeta)^Td\zeta}_\text{E}\times\dot{x_{1d}} = \left(\int_{0}^{L}\phi(\zeta)\ddot{\phi}(\zeta)d\zeta\right)e_{2d}
\end{equation}

On applique plusieurs fois de l’intégration par partie (IPP) :

\begin{equation*}
	\begin{aligned} 
		\int_{0}^{L}\phi(\zeta)\ddot{\phi}(\zeta)^T d\zeta &= [\phi(\zeta)\dot{\phi}(\zeta)^T]_{0}^{L} - \int_{0}^{L}\dot{\phi}(\zeta)\dot{\phi}(\zeta)^T d\zeta \\
		&= \phi(L)\dot{\phi}(L)^T - \underbrace{\phi(0)\dot{\phi}(0)^T }_\text{=0} - \dot{\phi}(L)\phi(L)^T + \underbrace{\dot{\phi(0)}\phi(0)^T }_\text{=0} + \int_{0}^{L} \ddot{\phi}(\zeta)\phi(\zeta)^T d\zeta \\
		&= \phi(L)\dot{\phi}(L)^T - \dot{\phi}(L)\phi(L)^T + \int_{0}^{L} \ddot{\phi}(\zeta)\phi(\zeta)^T d\zeta = D
	\end{aligned}
\end{equation*}

On continue avec ça en remplaçant dans l'équation 2 : $\Rightarrow \boxed{E\dot{x}_{1d} = De_{2d}}$ \\


Puis on a dans la deuxième ligne de l'équation 1 : 

\begin{equation*}
	\begin{aligned}
		\phi^T(\zeta)\dot{x}_{2d} &= - \frac{\partial^2}{\partial\zeta^2} (\phi(\zeta)^T e_{1d}) - q(\zeta, t) \\
		\int_{0}^{L} \phi^T(\zeta)d\zeta\times\dot{x}_{2d} &= -e_{1d}\int_{0}^{L}\phi(\zeta)\ddot{\phi}(\zeta)^T d\zeta - \underbrace{\int_{0}^{L}\phi(\zeta)q(\zeta,t)d\zeta}_{=F_{ext}}
	\end{aligned}
\end{equation*}

Puis on trouve D :

\begin{equation*}
	\begin{aligned}
		& D = \phi(L)\dot{\phi}(L)^T - \phi(L)\dot{\phi}(L)^T + \int_{0}^{L}\ddot{\phi}(\zeta)\phi(\zeta)^Td\zeta \\
		\Rightarrow & D^T = \dot{\phi}(L)\phi(L)^T - \dot{\phi}(L)\phi(L)^T + \int_{0}^{L}\phi(\zeta)\ddot{\phi}(\zeta)d\zeta \\
		\Rightarrow & \boxed{\int_{0}^{L}\underbrace{\phi(\zeta)}\ddot{\phi}(\zeta)^Td\zeta} = D^T - \dot{\phi}(L)\phi(L)^T + \phi(L)\dot{\phi}(L)^T \\
		\Rightarrow & E\dot{x}_{2d} = -e_{1d}D^T + \overbrace{e_{1d}\dot{\phi}(L)\phi(L)^T}^{=0} - e_{1d}\phi(L)\dot{\phi}(L)^T - F_{ext}
	\end{aligned}
\end{equation*}

Or on sait que : 

\begin{align*}
	\Rightarrow\left\{\begin{aligned}
		\frac{de_1}{d\zeta} = u(t) &\Rightarrow \dot{\phi}(L)^T e_{1d} = u(t) \\
		e_1 (L,t) = 0 &\Rightarrow \phi(L)^T e_{1d} = 0
	\end{aligned}\right. & &
	\Rightarrow & &
	\boxed{E\dot{x}_{2d} = - e_{1d}D^T - \phi(L)u(t) - F_{ext}}
\end{align*}

Puis on a ceci :

\begin{align*}
	\left\{\begin{aligned}
		y(t) &= -e_2 (L,t) \\
		e_2 (L,t) &\approx	\phi(L)^T e_{2d} 
	\end{aligned}\right. & &
	\Rightarrow & &
	\boxed{y(t) = -\phi(L)^T e_{2d}}
\end{align*}


\subsection{Question 3}
On a :

\begin{align*}
	\left\{\begin{aligned}
		e_1 = EIx_1 = x_1 &\Rightarrow e_{1d} = x_{1d} \\
		e_2 = \frac{1}{\rho(\zeta)}x_2 = x_2 &\Rightarrow e_{2d} = x_{2d}
	\end{aligned}\right. & &
	\Rightarrow & &
	\left\{\begin{aligned}
		E\dot{X}_{1d} &= Dx_{2d} \\
		E\dot{X}_{2d} &= -D^T x_{1d} - \phi(L)u(t) - 0 \\
		y &= -\phi(L)^T x_{2d}
	\end{aligned}\right.
\end{align*}

Ce qui nous donne :

\begin{equation*}
	A = \begin{pmatrix}
		0 & E^{-1}D \\
		-E^{-1}D^T & 0
	\end{pmatrix}, \  B = \begin{pmatrix}
		0 \\
		-E^{-1}\phi(L)
	\end{pmatrix}, \ C = \begin{pmatrix}
		0 & -\phi(L)^T
	\end{pmatrix},\ pour \ \left\{\begin{aligned}
	\zeta = L = 1 \\
	\phi(1) = \begin{pmatrix}
		0 \\
		1 \\
		0 \\
		0
	\end{pmatrix}
	\end{aligned}\right. 
\end{equation*}

Puis application numérique.