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So consider the heat equation on a rod of length $L$,

$u_t (x,t) = c^2 u_{xx} (x,t)$, $\forall (x,t) \in [0,L]$ x $\mathbb{R}^+ $,

and the energy at time $t$ defined as,

$$E(t)=\frac{1}{2}\int_{0}^{L} u(x,t)^2 dx.$$

How would I show that $E(t) \geq 0$ for every $t \in \mathbb{R}^+$, and that

$$ E'(t) = -c^2 \int_{0}^{L} (u_x (x,t))^2 dx + c^2 \big(u(L,t)u_x(L,t) - u(0,t)u_x(0,t)\big)? $$

Here's my attempt:

$E'(t) = \frac{d}{dt} \int_{0}^{L} \frac{u^2}{2} dx = \int_{0}^{L} \frac{1}{2} (u^2) dx = \int_{0}^{L} uu_t dx$

and if $u_t(x,t) = c^2 u_{xx}(x,t)$, then,

$E'(t) = c^2 \int_{0}^{L} u u_{xx} dx = \int_{0}^{L} uu_t dx$

But I don't really know where to go from here.

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Looking pretty good, though I'm not sure where your $u^2$ came from - the one multiplied by $u_{xx}$ in your last line. Now try an integration by parts. – Mark McClure Jan 28 '14 at 23:11
@MarkMcClure whoops, the $u^2$ is a typo! Thanks for catching that (i'll go ahead and edit that out). I completely forgot about integration by parts - haven't had to do it for awhile. Thanks for the tip, i'll try it out! – Lame-Ov2.0 Jan 28 '14 at 23:32
up vote 3 down vote accepted

Hint. Using integration by parts we obtain that $$ \int_0^L u(x,t)u_{xx}(x,t)\,dx=u(x,t)u_{x}(x,t)\,\big|_0^L-\int_0^L u_{x}^2(x,t)\,dx. $$

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Thanks! Yeah, I totally forgot about integration by parts (what @MarkMcClure suggested). Thanks for the verification – Lame-Ov2.0 Jan 29 '14 at 1:57

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