Toronto Math Forum
APM3462018S => APM346Tests => Term Test 1 => Topic started by: Victor Ivrii on February 15, 2018, 08:23:58 PM

Consider the PDE with boundary conditions:
\begin{align}
&u_{tt}+c^2u_{xxxx} + a u=0,\qquad&&0<x<L, \tag{1}\\
&u_{x=0} =u_{x}_{x=0}=0,\tag{2}\\
&u_{xx}_{x=L} =u_{xxx}_{x=0}=0_{x=L}=0, tag{3}
\end{align}
where $c>0$ and $a$ 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_{xx}^2 +au^2)\,dx
\tag{4}
\end{equation}
does not depend on $t$.

First take $\frac{dE}{dt} $ of (4):
We get $$ \int_{0}^{L} (u_t u_{tt} + c^2 u_{xx}u_{xxt} + a u u_t ) dx $$
Now, integrate the middle term by parts, twice. The nice thing is that due to those boundary conditions, we can simply write:
$$ \int_{0}^{L} (u_t u_{tt}  c^2 u_{xxx}u_{xt} + a u u_t ) dx + c^2u_{xx}u_{xt}_{0}^{L} $$
The last term evaluates to 0 at both 0 and L, so redoing this integration by parts gives us:
$$ \int_{0}^{L} (u_t u_{tt} + c^2 u_{xxxx}u_{t} + a u u_t ) dx + c^2u_{xxx}u_t_{0}^{L} $$
Which again allows for one of the terms $u_{xx} $ and $u_{xxx} $ to be evaluated as 0 at 0 and L.
$$ \int_{0}^{L} u_t ( u_{tt} + c^2 u_{xxxx} + a u ) dx $$
The factored term inside is clearly the PDE we set to be 0 (eqn 1).
Therefore our integrand is 0, and thus
$$ \frac{dE}{dt} = 0 $$
Therefore energy is conserved.