### Author Topic: P4 Night  (Read 880 times)

#### Victor Ivrii

• Elder Member
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##### P4 Night
« on: February 15, 2018, 08:23:58 PM »
Consider the PDE  with boundary conditions:
\begin{align}
&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

E(t)= \frac{1}{2}\int_0^L \bigl( u_t^2 + c^2u_{xx}^2 +au^2)\,dx
\tag{4}

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.