Author Topic: TT1 = Problem 3  (Read 6823 times)

Victor Ivrii

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TT1 = Problem 3
« on: October 16, 2012, 06:28:21 PM »
Consider the PDE  with boundary conditions:
\begin{align*}
&u_{tt}+K u_{xxxx} + \omega^2 u =0,\qquad&&0<x<L,\\[3pt]
&u(0,t)=u_x(0,t)=0,\\[3pt]
&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 $t$.
« Last Edit: October 16, 2012, 06:54:14 PM by Victor Ivrii »

Ian Kivlichan

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Re: TT1 = Problem 3
« Reply #1 on: October 16, 2012, 07:32:55 PM »
Solution to Question 3!

Re-wrote solution more nicely at Prof. Ivrii's request. Original at http://i.imgur.com/l4Pw2.jpg
« Last Edit: October 17, 2012, 01:08:18 AM by Ian Kivlichan »

Aida Razi

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Re: TT1 = Problem 3
« Reply #2 on: October 16, 2012, 07:39:46 PM »
Solution is attached,

Ian Kivlichan

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Re: TT1 = Problem 3
« Reply #3 on: October 16, 2012, 07:51:52 PM »
Solution is attached,

Aida: I'm not sure your solution is correct: u(0,t)=0 and u_x(0,t)=0 don't necessarily imply that u_xx(0,t) = 0. Consider for example u(x, t) = x^2. There, u_xx(0, t) = 2 despite u(0,t)=0 and u_x(0,t)=0.

Up until crossing out u_xx on the last line, though, I think your solution is still right, and your final answer is definitely right. ;P

Victor Ivrii

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Re: TT1 = Problem 3
« Reply #4 on: October 16, 2012, 09:54:02 PM »
Aida, your conclusion that out of integral terms vanish is correct but not a justification. Ian is correct both in the solution and finding flaws in Aida' solution. Sure these terms disappear as bot $u_t$ and $u_{tx}$ vanish as $x=0$ and $x=L$ due to boundary condition.

Remark. With $\gamma=0$ this equation
\begin{equation}
u_{tt}+Ku_{xxxx}+\gamma u=0
\end{equation}
 describes oscillations of the beam (beam resists to bending rather than stretching; in fact resistance to bending is due to the fact that while bend the outer part of the beam (which is on its convex side) stretches and inner part of the beam (which is on its concave side) squeezes.

Picture attached taken from

http://www.flickr.com/photos/mitopencourseware/3364718263/



With $\gamma<0$ this equation describes oscillation of the rotating shaft. We have $\gamma=\omega^2$ which is less interesting but makes $E$ positively definite.

PS. Ian, your posts are virtually useless for a class: too poor handwriting makes it almost impossible to read for anyone who does not know solution. Could you repost?
PPS. No need--Zarak did it nicely, and you did it too :D
« Last Edit: October 17, 2012, 01:53:22 AM by Victor Ivrii »

Zarak Mahmud

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Re: TT1 = Problem 3
« Reply #5 on: October 16, 2012, 10:31:29 PM »
Here is a rather brief typed version.

\begin{equation} E(t)= \frac{1}{2}\int_0^L \big( u_t^2 + K u_{xx}^2 + \omega^2 u^2\big)\,dx \end{equation}

Take the derivative with respect to $t$ and then integrate the middle term by parts.

\begin{equation*} E'(t)= \frac{1}{2}\int_0^L \bigl( u_{tt}u_t + K u_{xx}u_{xxt} + \omega^2 uu_t\big)\,dx \\

K \left( u_{xt}u_{xx}\big|_{0}^{L} - \int_0^L u_{xt}u_{xxx}dx \right) \\
\implies K \left( u_{xt}u_{xx}\big|_{0}^{L} - u_{t}u_{xxx}\big|_{0}^{L} + \int_0^L u_tu_{xxxx} \right)\\

E'(t)= \int_0^L u_t \bigl( u_{tt}+K u_{xxxx} + \omega^2 u \big) dx +  K \big[ u_{xt}u_{xx} - u_{t}u_{xxx} \big]_{0}^{L}\\

=K \big[ u_{xt}u_{xx} - u_{t}u_{xxx} \big]_{0}^{L}

\end{equation*}

After plugging in the bounds and checking boundary conditions, we see that $E'(t) = 0$ and thus $E(t)$ is a constant function, independent of $t$.
« Last Edit: October 19, 2012, 01:38:29 PM by Zarak Mahmud »

Ian Kivlichan

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Re: TT1 = Problem 3
« Reply #6 on: October 17, 2012, 01:11:39 AM »
PS. Ian, your posts are virtually useless for a class: too poor handwriting makes it almost impossible to read for anyone who does not know solution. Could you repost?

Sorry!!  :(

I have tried my best to re-write it nicely (edited original post).