\textbf{Question 2}

\begin{equation}
\dot{x}_1
=
\frac{\partial^2}{\partial \zeta^2} e_2(\zeta,t)
\end{equation}

Par la méthode de discrétisation :

\begin{equation}
\phi^T(\zeta)\dot{x}_{1d}(t)
=
\frac{\partial^2 \phi^T(\zeta)}{\partial \zeta^2} e_{2d}(t) \Leftrightarrow \int_0^L \phi(\zeta)\phi^T(\zeta)\,d\zeta \; \dot{x}_{1d}(t)
=
\underbrace{
\int_0^L
\phi(\zeta)
\frac{\partial^2 \phi^T(\zeta)}{\partial \zeta^2}
\,d\zeta
e_{2d}(t)
}_{J}
\end{equation}

Par intégration par parties deux fois :

\begin{equation}
J
=
\left[
\phi(\zeta)
\frac{\partial \phi^T(\zeta)}{\partial \zeta}
\right]_0^L
-
\int_0^L
\frac{\partial \phi(\zeta)}{\partial \zeta}
\frac{\partial \phi^T(\zeta)}{\partial \zeta}
\,d\zeta
\end{equation}

\begin{equation}
\begin{aligned}
J 
&=
\phi(L)\frac{\partial \phi^T}{\partial \zeta}(L)e_{2d}(t)
-0
-
\left[
\frac{\partial \phi}{\partial \zeta}(\zeta)
\phi^T(\zeta)e_{2d}(t)
\right]_0^L
+
\int_0^L
\frac{\partial^2 \phi}{\partial \zeta^2}(\zeta)
\phi^T(\zeta)\,d\zeta \; e_{2d}(t)
\\[0.2cm]
&=
\phi(L)\frac{\partial \phi^T}{\partial \zeta}(L)e_{2d}(t)
-
\frac{\partial \phi}{\partial \zeta}(L)
\phi^T(L)e_{2d}(t)
+
\frac{\partial \phi}{\partial \zeta}(0)
e_{2d}(0,t)
+
\int_0^L
\frac{\partial^2 \phi}{\partial \zeta^2}(\zeta)
\phi^T(\zeta)\,e_{2d}(t)d\zeta \; 
\end{aligned}
\end{equation}


On a $e_2(0,t)=0 \Leftrightarrow E\dot{x}_1 = De_{2d}(t)$


Comme $D =
\phi(L)\frac{d\phi^T}{d\zeta}(L)
-
\frac{d\phi}{d\zeta}(L)\phi^T(L)
+
\int_0^L
\frac{d^2\phi}{d\zeta^2}(\zeta)\phi^T(\zeta)\,d\zeta
$. Et $(AB)^T = B^T A^T$, on a :

\begin{equation}
-D^T =
-
\frac{d\phi}{d\zeta}(L)\phi^T(L)
+
\phi(L)\frac{d\phi^T}{d\zeta}(L)
-
\int_0^L
\phi(\zeta)
\frac{d^2\phi^T}{d\zeta^2}(\zeta)
\,d\zeta.
\end{equation}

\begin{equation}
E\dot{x}_{2d}(t)
=
-
\int_0^L
\phi(\zeta)
\frac{d^2\phi^T}{d\zeta^2}(\zeta)
\,d\zeta \; e_{1d}(t)
-
F_{ext}.
\end{equation}

Donc, 
\begin{equation}
E\dot{x}_{2d}(t)
=
-D^T e_{1d}(t)
+
\frac{d\phi}{d\zeta}(L)e_1(L,t)
-
\phi(L)\frac{\partial e_1}{\partial \zeta}(L,t)
-
F_{ext}.
\end{equation}

Comme $e_1(L,t)=0$, $\frac{\partial e_1}{\partial \zeta}(L,t)=u(t)$ on obtient bien :

\begin{equation}
E\dot{x}_{2d}(t)
=
-D^T e_{1d}(t)
-
\phi(L)u(t)
-
F_{ext}.
\end{equation}