# Toronto Math Forum

## APM346-2016F => APM346--Tests => TT1 => Topic started by: Victor Ivrii on October 19, 2016, 10:29:21 PM

Title: TT1-P4
Post by: Victor Ivrii on October 19, 2016, 10:29:21 PM
Consider the PDE  with boundary conditions:
\begin{align}
&(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

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
does not depend on $t$.
Title: Re: TT1-P4
Post by: Roro Sihui Yap 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 $\partial_x(u_tu_x) = u_{xt}u_x + u_tu_{xx}$
$$\partial_t E(t) = c^2(u_tu_x)\big|_{0}^{L} + c^2 u_t(0,t)u_x(0,t) - c^2 u_t(L,t)u_x(L,t)$$
$$\partial_t E(t) = c^2u_t(L,t)u_x(L,t) - c^2u_t(0,t)u_x(0,t) + c^2 u_t(0,t)u_x(0,t) - c^2 u_t(L,t)u_x(L,t) =0$$
Title: Re: TT1-P4
Post by: Victor Ivrii on October 20, 2016, 04:52:52 AM
:D