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\newpage
\pagenumbering{Alph}
\section{Annexe}
\vspace*{-1em}
\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 1\textsuperscript{ère} ligne de l'éq. 1, on trouve :
\vspace*{-.2em}
\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) :
\vspace*{-.5em}
\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*}
\vspace*{-.5em}
Continuons en remplaçant dans l'équation 2 : $\Rightarrow \boxed{E\dot{x}_{1d} = De_{2d}}$ Dans la 2\textsuperscript{de} ligne de l'éq. 1 on a : 

\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*}
\vspace*{-1.3em}

Puis on trouve D :
\vspace*{-.5em}
\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*}
\vspace*{-.8em}
\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 :
\vspace*{-.1em}
\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}
\vspace*{-2em}
\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*}
\vspace*{-.5em}
\begin{equation*}
	\Rightarrow 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*}
\vspace*{-1.5em}\\
Puis application numérique
\vspace*{-2em}\\
\subsection{Question 4}
\vspace*{-2em}
\begin{equation*}
	\omega(p, t) = C_\omega(p)x_d(t) = \begin{bmatrix}
		\frac{p^2(2p^3-5p^2+10)}{20} & -\frac{p^4(2p-5)}{20} & \frac{p^3(3p^2-10p+10)}{60} & \frac{p^4(3p-5)}{60} & 0 & 0 & 0 & 0
	\end{bmatrix}x_d(t)
\end{equation*}

\begin{equation*}
	\phi(L)^T = \begin{bmatrix}
		2p^3 - 3p^2 + 1 & 3p^2 - 2p^3 & p^3 - 2p^2 + p & p^3 - p^2
	\end{bmatrix}
\end{equation*}
\vspace*{-4em}\\
\subsection{Question 5}
\vspace*{-2em}
\begin{align*}
	y  = -\phi(L)^Te_{2d} && y = -e_2(L, t) & & e_2(\zeta, t) \approx \phi(\zeta)^Te_{2d}(t)
\end{align*}
\vspace*{-2em}
\begin{align*}
	x_1 = \frac{\partial^2\omega}{\partial\zeta^2}(\zeta,t)\approx\phi(\zeta)^2x_{1d} & & x_2(\zeta,t) = \rho(\zeta)\frac{\partial\omega}{\partial t}(\zeta,t) \approx\phi(\zeta)^Tx_{2d}
\end{align*}
\vspace*{-2em}
\begin{align*}
	\begin{bmatrix}
		\dot{x}_{1d} \\ \dot{x}_{1d}
	\end{bmatrix} = A \begin{bmatrix}
		x_{1d} \\ x_{2d}
	\end{bmatrix} + Bu & & y = C \begin{bmatrix}
		x_{1d} \\ x_{2d}
	\end{bmatrix}
\end{align*}
\vspace*{-4em}\\

\subsection{Question 7}

\vspace*{-.5em}
La sortie est donnée par $y(t) = C\,x_d(t)$, donc le système en boucle fermée s’écrit :

\begin{equation*}
	\begin{cases}
		\dot{x}_d(t) = (A - B C k)\,x_d(t) + B H\,w_c(L,t) \\
		w(L,t) = C_w(L)\,x_d(t)
	\end{cases}
\end{equation*}

En régime permanent, $\dot{x}_d(t)=0$, donc :
\begin{equation*}
	0 = (A - B C k)\,x_d + B H\,C_w(L)\,x_d
	\quad \Rightarrow \quad
	x_d = -(A - B C k)^{-1} B H w_c(L,t)
\end{equation*}

En multipliant l’expression par $C_w(L)$, on trouve :
\begin{equation*}
	w(L,t) = C_w(L)\,x_d = -C_w(L)(A - B C k)^{-1} B H w_c(L,t)
\end{equation*}

Pour assurer le suivi $w(L,t) = w_c(L,t)$, il faut :
\begin{equation*}
	- C_w(L)(A - B C k)^{-1} B H = I
\end{equation*}
\vspace*{-4em}\\

\subsection{Question 15}
\vspace*{-.5em}
On sait que :
\begin{equation*}
	F_{ext} = \int_{0}^{L}\phi(\zeta)q(\zeta,t)d\zeta
	\quad \Rightarrow \quad 
	F_{ext} = q_0(t) \cdot \int_{0}^{L}\phi(\zeta)d\zeta\ \text{ car }\ q(\zeta,t) = q_0(t)
\end{equation*}
Or on a $E\dot{x}_{2d} = - e_{1d}D^T - \phi(L)u(t) - F_{ext}$\\
Donc, puisque $\dot{x}_{d}(t) = A x_d(t) + B u(t) + B_p q_0(t)$, on trouve les résultats d'en-haut.