134 lines
2.3 KiB
TeX
134 lines
2.3 KiB
TeX
\textbf{Question 2}
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\begin{equation}
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\dot{x}_1
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=
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\frac{\partial^2}{\partial \zeta^2} e_2(\zeta,t)
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\end{equation}
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Par la méthode de discrétisation :
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\begin{equation}
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\phi^T(\zeta)\dot{x}_{1d}(t)
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=
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\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)
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=
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\underbrace{
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\int_0^L
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\phi(\zeta)
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\frac{\partial^2 \phi^T(\zeta)}{\partial \zeta^2}
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\,d\zeta
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e_{2d}(t)
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}_{J}
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\end{equation}
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Par intégration par parties deux fois :
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\begin{equation}
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J
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=
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\left[
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\phi(\zeta)
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\frac{\partial \phi^T(\zeta)}{\partial \zeta}
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\right]_0^L
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-
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\int_0^L
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\frac{\partial \phi(\zeta)}{\partial \zeta}
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\frac{\partial \phi^T(\zeta)}{\partial \zeta}
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\,d\zeta
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\end{equation}
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\begin{equation}
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\begin{aligned}
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J
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&=
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\phi(L)\frac{\partial \phi^T}{\partial \zeta}(L)e_{2d}(t)
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-0
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-
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\left[
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\frac{\partial \phi}{\partial \zeta}(\zeta)
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\phi^T(\zeta)e_{2d}(t)
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\right]_0^L
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+
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\int_0^L
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\frac{\partial^2 \phi}{\partial \zeta^2}(\zeta)
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\phi^T(\zeta)\,d\zeta \; e_{2d}(t)
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\\[0.2cm]
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&=
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\phi(L)\frac{\partial \phi^T}{\partial \zeta}(L)e_{2d}(t)
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-
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\frac{\partial \phi}{\partial \zeta}(L)
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\phi^T(L)e_{2d}(t)
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+
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\frac{\partial \phi}{\partial \zeta}(0)
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e_{2d}(0,t)
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+
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\int_0^L
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\frac{\partial^2 \phi}{\partial \zeta^2}(\zeta)
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\phi^T(\zeta)\,e_{2d}(t)d\zeta \;
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\end{aligned}
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\end{equation}
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On a $e_2(0,t)=0 \Leftrightarrow E\dot{x}_1 = De_{2d}(t)$
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Comme $D =
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\phi(L)\frac{d\phi^T}{d\zeta}(L)
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-
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\frac{d\phi}{d\zeta}(L)\phi^T(L)
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+
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\int_0^L
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\frac{d^2\phi}{d\zeta^2}(\zeta)\phi^T(\zeta)\,d\zeta
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$. Et $(AB)^T = B^T A^T$, on a :
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\begin{equation}
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-D^T =
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-
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\frac{d\phi}{d\zeta}(L)\phi^T(L)
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+
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\phi(L)\frac{d\phi^T}{d\zeta}(L)
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-
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\int_0^L
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\phi(\zeta)
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\frac{d^2\phi^T}{d\zeta^2}(\zeta)
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\,d\zeta.
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\end{equation}
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\begin{equation}
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E\dot{x}_{2d}(t)
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=
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-
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\int_0^L
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\phi(\zeta)
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\frac{d^2\phi^T}{d\zeta^2}(\zeta)
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\,d\zeta \; e_{1d}(t)
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-
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F_{ext}.
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\end{equation}
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Donc,
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\begin{equation}
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E\dot{x}_{2d}(t)
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=
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-D^T e_{1d}(t)
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+
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\frac{d\phi}{d\zeta}(L)e_1(L,t)
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-
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\phi(L)\frac{\partial e_1}{\partial \zeta}(L,t)
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-
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F_{ext}.
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\end{equation}
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Comme $e_1(L,t)=0$, $\frac{\partial e_1}{\partial \zeta}(L,t)=u(t)$ on obtient bien :
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\begin{equation}
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E\dot{x}_{2d}(t)
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=
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-D^T e_{1d}(t)
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-
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\phi(L)u(t)
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-
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F_{ext}.
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\end{equation}
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