### Author Topic: Web bonus problem : Week 4 (#3)  (Read 2213 times)

#### Victor Ivrii

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##### Web bonus problem : Week 4 (#3)
« on: October 04, 2015, 05:51:30 AM »

#### Emily Deibert

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##### Re: Web bonus problem : Week 4 (#3)
« Reply #1 on: October 22, 2015, 12:13:29 PM »
I will start this one.

We take the time derivative of the energy as given in the problem. So:

\frac{1}{2}\int_0^{\infty}[(2u_tu_{tt} + 2ku_{xx}u_{xxt}]dx = \int_0^{\infty}[u_t(-ku_{xxxx}) + ku_{xx}u_{xxt}]dx

Now we will integrate by parts twice on the second term in this integral. I will omit the steps and state the result:

-k\int_0^{\infty}u_tu_{xxxx}dx +ku_{xx}u_{xt}|_0^{\infty} - ku_tu_{xxx}|_0^{\infty} + k\int_0^{\infty}u_tu_{xxxx}dx

The first and last terms cancel. Let's make use of the first boundary condition.

u|_{x=0} = u_x|{x=0} = 0 \longrightarrow u_t|_{x=0} = u_{xt}|_{x=0} = 0

We plug this in to get the final result:
\frac{\partial{}E(t)}{\partial{}t} = 0

I will add the other BCs later; or if someone wants to collaborate and add those ones please do!

#### Emily Deibert

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##### Re: Web bonus problem : Week 4 (#3)
« Reply #2 on: October 22, 2015, 12:18:56 PM »
Sorry, forgot to mention that we probably need to make the assumption that u is fast-decaying.

#### Victor Ivrii

Boundary terms are $u_t u_{xxx}-u_{tx}u_{xx}$; so if we assume that $u=u_x=0$ or $u=u_{xx}=0$ or $u_{xx}=u_{xxx}=0$ on the left/right end we get $0$ here