Author Topic: TT1-P4  (Read 251 times)

Victor Ivrii

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TT1-P4
« on: October 19, 2016, 10:29:21 PM »
Consider the PDE  with boundary conditions:
\begin{align}
&u_{tt}-c^2u_{xx}  =0,\qquad&&0<x<L,\label{eq-4-1}\\
&(u_x -\alpha u_{tt})|_{x=0}=0,\label{eq-4-2}\\
&(u_x +\beta u_{tt})|_{x=L}=0\label{eq-4-3}
\end{align}
where  $c>0$ and $\alpha>0$ are constant. Prove that the energy $E(t)$ defined as
\begin{equation}
E(t)= \frac{1}{2}\int_0^L \bigl( u_t^2 + c^2u_{x}^2 \bigr)\,dx +c^2\frac{\alpha}{2}u_t(0,t)^2+
c^2\frac{\beta}{2}u_t(L,t)^2\end{equation}
does not depend on $t$.

Roro Sihui Yap

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Re: TT1-P4
« Reply #1 on: October 19, 2016, 10:41:51 PM »
 Want to prove that $\partial_t E(t) = 0 $
$$\partial_t E(t) = \frac{1}{2}\int_0^L ( 2u_tu_{tt} + 2c^2u_{x}u_{xt} )\,dx + (c^2\alpha) u_t(0,t)u_{tt}(0,t)+  (c^2\beta) u_t(L,t)u_{tt}(L,t) $$

Use $u_{tt} = c^2u_{xx} $, $\alpha u_{tt}(0,t) = u_x(0,t) $ and $\beta u_{tt}(L,t) = -u_x(L,t)$
$$\partial_t E(t) = \int_0^L (c^2 u_tu_{xx} + c^2u_{x}u_{xt} )\,dx + c^2 u_t(0,t)u_x(0,t) - c^2 u_t(L,t)u_x(L,t) $$
Since $\partial_x(u_tu_x) = u_{xt}u_x + u_tu_{xx}$
$$\partial_t E(t) = c^2(u_tu_x)\big|_{0}^{L} + c^2 u_t(0,t)u_x(0,t) - c^2 u_t(L,t)u_x(L,t) $$
$$\partial_t E(t) = c^2u_t(L,t)u_x(L,t) - c^2u_t(0,t)u_x(0,t) + c^2 u_t(0,t)u_x(0,t) - c^2 u_t(L,t)u_x(L,t) =0  $$
« Last Edit: October 19, 2016, 10:58:52 PM by Roro Sihui Yap »

Victor Ivrii

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Re: TT1-P4
« Reply #2 on: October 20, 2016, 04:52:52 AM »
:D