### Author Topic: TT1-P4  (Read 2775 times)

#### Victor Ivrii

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##### TT1-P4
« on: October 21, 2015, 08:50:32 PM »
Consider the PDE  with boundary conditions:
\begin{align}
&u_{tt}-c^2u_{xx} + \omega^2 u =0,\qquad&&0<x<L,\label{eq-4-1}\\[2pt]
&(u_x -\alpha u)|_{x=0}=0,\label{eq-4-2}\\[2pt]
&(u_x +\beta u)|_{x=L}=0\label{eq-4-3}
\end{align}
where  $c>0$ and $\omega>0$ areconstant. Prove that the energy $E(t)$ defined as

E(t)= \frac{1}{2}\int_0^L \bigl( u_t^2 + c^2u_{x}^2 + \omega^2 u^2)\,dx +\frac{c^2\alpha}{2}u(0,t)^2+
\frac{c^2\beta}{2}u(L,t)^2

does not depend on $t$.

#### Yiqi Shi

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##### Re: TT1-P4
« Reply #1 on: October 21, 2015, 10:18:33 PM »

\frac{dE(t)}{dt}= \frac{1}{2}\int_0^L \bigl( 2u_tu_{tt}+ 2u_{x}c^2u_{xt} + \omega^22 uu_t)\,dx +c^2\alpha u(0,t)u_t(0,t)+ c^2\beta u(L,t)u_t(L,t)

do integral by part on $\int_0^L \bigl(2u_{x}c^2u_{xt})\,dx$

\int_0^L \bigl(2u_{x}c^2u_{xt})\,dx=2c^2[(u_{x}u_{t})\Big|_{0}^L - \int _0^L u_{xx}u_{t}\,dx)]

plugging these into $\frac{dE(t)}{dt}$

\frac{dE(t)}{dt}=c^2\int_0^L \bigl( u_tu_{tt}- u_{xx}u_{t} + \omega^2 uu_t)\,dx +c^2(u_{x}u_{t})\Big|_{0}^L  +c^2\alpha u(0,t)u_t(0,t)+ c^2\beta u(L,t)u_t(L,t)

By condition $u_{tt}-c^2u_{xx} + \omega^2 u =0$

c^2\int_0^L \bigl( u_tu_{tt}- u_{xx}u_{t} + \omega^2 uu_t)\,dx=0

\frac{dE(t)}{dt}=c^2(u_{x}u_{t})\Big|_{0}^L  +c^2\alpha u(0,t)u_t(0,t)+ c^2\beta u(L,t)u_t(L,t)

by condition $(u_x -\alpha u)|_{x=0}=0$ and $(u_x +\beta u)|_{x=L}=0$, we have

\begin{split}
&\alpha u(0,t)=u_{x}(0,t)\\
&\beta u(L,t)=-u_{x}(L,t)
\end{split}

plugging (6) into  $\frac{dE(t)}{dt}$, we have:

\begin{split}
\frac{dE(t)}{dt}&=c^2[(u_{x}u_{t})\Big|_{0}^L+u_{x}(0,t)u_t(0,t)-u_{x}(L,t)u_t(L,t)]\\
&=c^2[u_{x}(L,t)u_t(L,t)-u_{x}(0,t)u_t(0,t)+u_{x}(0,t)u_t(0,t)-u_{x}(L,t)u_t(L,t)]\\
&=c^2[0+0]\\
&=0
\end{split}

Since $\frac{dE(t)}{dt}$=0, then E(t) does not depend on t.