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Show that if u solves the KDV equation

$u_t + u_{xxx} + 6uu_x = 0$ for $x \in \mathbb{R}$, $t > 0$

then the energy

$\int_{-\infty}^{\infty} \frac{1}{2} u_x ^2 – u^3 \,dx$

is constant in time.

Attempt: The usual idea is to differentiate under the integral and then maybe do integration by parts, but I couldn’t find anything nice. Any hints? Thanks in advance.

EDIT:

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Source: S06_Final_Exam-L.Evans.pdf

Actual Attempt: Put $e(t) = \int_{-\infty}^{\infty} \frac{1}{2} u_x ^2 – u^3 \,dx$. Differentiating under the integral and then integrating by parts, we have $e'(t) = \int_{-\infty}^{\infty} u_x u_{xt} – 3u^2 u_t \,dx = u_x u_t |^{+\infty}_{-\infty} – \int_{-\infty}^{\infty} u_{xx} u_t \,dx – \int_{\infty}^{\infty} 3u^2 u_t \,dx$.

Assuming u has compact support, the first term vanishes, so we’re left with

$e'(t) = -\int_{-\infty}^{\infty}u_t(u_{xx} + 3u^2)\,dx$.

The only reason why I think this may be useful is the fact that we can rewrite the KDV equation as $u_t + (u_{xx})_x + (3u^2)_x = 0$ and integrate. But I don’t see how to finish the problem from here, which is why I’m looking for other suggestions.

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So define $$ W = u_{xx} + 3u^2, $$ your expression

$u_t + (u_{xx})_x + (3u^2)_x = 0$ says

$u_t = – W_x. $

Then

$$e'(t) = \int_{-\infty}^{\infty} \; W_x \, W \,dx,$$ and

$$e'(t) = \frac{1}{2} W^2 \left|^{x=+\infty}_{x=-\infty} \right. $$

So if $u,$ as a function of $x,$ has compact support or is, for example, in the Schwartz class, then your $e'(t) = 0.$

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