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I calculated the

$$ \begin{align} \frac{dE(t)}{2\,dt} & = \int_\Omega u_tu_{tt}+DuDu_t+u^3u_t\,dx \\ & =\int_\Omega [u_t(u_{tt}-\Delta u)+u^3u_t] \, dx+\int_{\partial \Omega} u_t \frac{\partial u}{\partial v} \, ds = \int_{\partial \Omega} u_t \frac{\partial u}{\partial v} \, ds \end{align} $$

However, I don't know how to prove $$\int_{\partial \Omega} u_t \frac{\partial u}{\partial v}ds$$ is zero by the conditions given in the problem?

Can anyone help me about this? Thanks so much!

Can anyone help me explain why $u_t\cong 0$ on $\partial \Omega$? Thanks so much!:)

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Notice $u_t\cong 0$ on $\partial \Omega$!

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  • $\begingroup$ Thanks! But can you explain for me why that is $0$? I cannot see that from the problem. Thanks again! $\endgroup$ – Sherry Dec 16 '14 at 1:26
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    $\begingroup$ $u(x,t) = 0$ in $\partial \Omega \times [0,T]$ so.. $\endgroup$ – JessicaK Dec 16 '14 at 1:43
  • $\begingroup$ It might help you to look at it this way: fix a $x_0 \in \partial \Omega$ and consider the function $g(t) = u(x_0,t)$. What is the value of $g$ for all $t$? What is the derivative of such a function? Was anything special about $x_0$? :) $\endgroup$ – Jason Knapp Dec 16 '14 at 1:44

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