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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:

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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  • $\begingroup$ If $u$ is a nonzero constant, KDV is satisfied but your integral is not defined. $\endgroup$ – Will Jagy May 11 '11 at 2:47
  • $\begingroup$ Good point. I took this problem verbatim from a final exam given by Lawrence Evans, so I think some conditions are implicit. So let's suppose u has compact support; in particular we rule out the trivial solution u = 0. Does the proposition still hold? $\endgroup$ – David May 11 '11 at 3:42
  • $\begingroup$ Final exam for what course? $\endgroup$ – Will Jagy May 11 '11 at 4:00
  • $\begingroup$ Math 126. I don't go to Berkeley but I am working through Evans' PDE text. $\endgroup$ – David May 11 '11 at 4:06
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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 $e'(t) = 0.$

http://en.wikipedia.org/wiki/Schwartz_space

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