# KDV PDE: energy constant in time

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.

• If $u$ is a nonzero constant, KDV is satisfied but your integral is not defined. – Will Jagy May 11 '11 at 2:47
• 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? – David May 11 '11 at 3:42
• Final exam for what course? – Will Jagy May 11 '11 at 4:00
• Math 126. I don't go to Berkeley but I am working through Evans' PDE text. – David May 11 '11 at 4:06

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 $e'(t) = 0.$