TT1 Problem 3

[Victor Ivrii]

Consider the PDE  with boundary conditions:
\begin{align*}
&u_{tt}+K u_{xxxx} + \omega^2 u =0,\qquad&&0<x<L,\\
&u_{xx} (0,t)=u_{xxx}(0,t)=0,\\
&u(L,t)=u_x(L,t)=0,
\end{align*}
where  $K>0$ is constant. Prove that the energy $E(t)$ defined as
\begin{equation*}
E(t)= \frac{1}{2}\int_0^L \bigl( u_t^2 + K u_{xx}^2 + \omega^2 u^2)\,dx
\end{equation*}
does not depend on $t$.

[Ping Wei]

Solution to Question 3

[Biao Zhang]

\begin{align*}
\frac{dE(t)}{dt}=\frac{1}{2}\int _0^L \frac{d}{dt}(u_t^2+Ku_{xx}^2+\omega^2u^2)\,dx\\
=\frac{1}{2}\int _0^L (2u_tu_{tt}+2Ku_{xx}u_{xxt}+2\omega^2uu_t)\,dx\\
=\int _0^L (u_tu_{tt}+Ku_{xx}u_{xxt}+\omega^2uu_t)\,dx\\
\end{align*}
Integrate $K\int _0^L u_{xx}u_{xxt}\,dx$ by part.\\
let $a=u_{xx}$, $db=u_{xxt}dx$. we have $da=u_{xxx}dx$,$b=u_{xt}$, then\\
\begin{equation*}
K((u_{xx}u_{xt})\Big|_{0}^L - \int _0^L u_{xxx}u_{xt}\,dx) \\
\end{equation*}
Integrate by part again, let $a'=u_{xxx}$,$db'=u_{xx}dx$. We have $da'=u_{xxxx}dx$ and $b'=u_t$. Then:
\begin{align*}
\rightarrow K(u_{xx}u_{xt})\Big|_{0}^L-K(u_tu_{xxx})\Big|_{0}^L+\int _0^L Ku_{xxxx}u_t\, dx\\
\rightarrow \int _0^L u_t\underbrace{(u_{tt}+Ku_{xxxx}+\omega^2u)}_0 dx+K(u_{xx}u_{xt}-u_tu_{xxx})\Big|_0^L\\
=K(u_{xx}u_{xt}-u_tu_{xxx})\Big|_0^L
\end{align*}
Since $u_{xx}(0,t)=u_{xxx}(0,t)=u(L,t)=u_x(L,t)=0$\\
We find out that $\frac{dE(t)}{dt}=0$, therefore $E(t)$ is not depend on $t$.

[Victor Ivrii]

OK, Biao
Ping, please do not attach pdfs  especially produced by a scanner. png or jpeg would be preferable.

It is important that in  the solution written
* integration by parts is done and
* the out-of integral terms are written and explained why these terms are $0$.