Why is the energy of Korteweg-deVries equation constant in time? The Korteweg-deVries is $u_t+6uu_x+u_{xxx}=0$ I know that its energy is 
$$E(t)=\int_{-\infty}^\infty (\frac12(u_x)^2-u^3)dx$$.
I know $u(x,t)$ and $u'(x,t)$ decays to 0 as $x\rightarrow \pm\infty$. If $E(t)$ is constant w.r.t. $t$ is it sufficient to only show $E'(t)=0$?

This isn't a complete answer but it gives some insight into why the energy is constant for wave solutions, $u(x,t) = f(x - ct)$.
Plugging this into $u_t+6uu_x+u_{xxx}=0$ reduces to a third-order ODE, $-cf' +6ff' + f'''=0$, which can be integrated to $-cf + 3f^2 + f''= C_1$. 
If you multiply by $f'$ and integrate a second time you get a constant of motion: 
$$-\dfrac{1}{2} c f^2 + f^3 + \dfrac{1}{2}(f')^2 - C_1 f = E$$
This is conservation of energy for the KdV equation with wave solutions.
