Invariants of KdV equation (Non-linear PDE)

[McCoy13]

Homework Statement

Show that the mass is a constant of the motion (invariant) for the KdV equation by direct differentiation with respect to time.

Homework Equations

KdV equation: $u_{t}+u_{xxx}+6uu_{x}=0$
mass: $\int udx$
(integral is taken over whole line)

The Attempt at a Solution

$$\frac{d}{dt} \int u dx$$

[tex]\int u_{t} dx[/itex]

$$-\int(6uu_{x}+u_{xxx})dx$$

The first term in the integral integrates to 3u^2 which is 0 when integrated over the whole line by even symmetry. I am unsure what to do with the $u_{xxx}$ term.

I also suspect I may be doing the differentiation (if x is supposed to be parametrized by t or something, I dunno) and/or integration wrong, but I may also just not see the solution to the right hand term.

 

[mighty2000]

First, let's rewrite the KdV equation in a more compact form:

u_t + u_{xxx} + 6uu_x = 0

Now, let's differentiate the mass integral with respect to time:

\frac{d}{dt} \int udx = \int u_t dx

Next, we can use the KdV equation to substitute for u_t and u_{xxx}:

\int (-6uu_x)dx = -6\int u u_x dx

Now, we can use integration by parts to solve the right hand side of the equation:

\int u u_x dx = \frac{1}{2}u^2 - \int u_x^2 dx

Since we are integrating over the whole line, the first term on the right hand side will be 0 due to even symmetry. This leaves us with:

\frac{d}{dt} \int udx = -6\int u_x^2 dx

Since the integral on the right hand side is always non-negative, this shows that the mass is a constant of motion (invariant) for the KdV equation.