Author Topic: P4 Night  (Read 147 times)

Victor Ivrii

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P4 Night
« 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 $t$.

Tristan Fraser

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Re: P4 Night
« Reply #1 on: February 16, 2018, 03:24:59 PM »
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.