Conserved quantities in the Korteweg-de Vries equation

[Emil_M]

Homework Statement

Consider the Kortweg-de Vires Equation in the form

$$\frac{\partial \psi}{\partial t}+\frac{\partial^3 \psi}{\partial x^3}+6\psi\frac{\partial \psi}{\partial x}=0$$

Find the relation between the coefficients $c$ and $d$ , such that the following quantity is conserved:

$$c\; \int_{-\infty}^\infty\left(\frac{\partial\psi}{\partial x}\right)^2 \mathrm{d}x+d\;\int_{-\infty}^{\infty}\psi ^3 \mathrm{d}x$$

Homework Equations

A local conservation law is of the form $D_t+F_x=0$.

The Attempt at a Solution

Usually, I would try to create a local conservation law, s.t. the quantity in question is conserved. But I really don't know how to do this in the given case. Thanks!

 

[Emil_M]

Hey, I've attempted to do the following:

Since the quantity in question needs to be conserved, it follows that

$$\frac{\partial}{\partial t}\left( \; \int_{-\infty}^\infty c\left(\frac{\partial\psi}{\partial x}\right)^2 \mathrm{d}x+d\;\int_{-\infty}^{\infty}\psi ^3 \mathrm{d}x \right) =0$$.

Thus, I have computed the temporal derivation, which yields

$$-c\int_{-\infty}^\infty \mathrm{d}x \left( 2c\frac{\partial^4 \psi}{\partial x^4}\frac{\partial\psi}{\partial x}+12c\left(\frac{\partial \psi}{\partial x}\right)^2+12c \psi\frac{\partial\psi}{\partial x}\frac{\partial^2 \psi}{\partial x^2}+3d\psi^2\frac{\partial^3\psi}{\partial x^3}+18\psi ^3 \frac{\partial \psi}{\partial x} \right)=0 $$

if I define $d:=-c$.

If I can show that the integrand is a perfect derivative, I am finished. However, I can't seem to show that this is the case. Can anyone help me?

 

[John Park]

I don't think you should have the factor of c outside the integral, but you need a factor of d in the last term of the integrand. (But I'm also one of those that deplore using d as a constant in anything involving calculus.)

It looks as though the integrand should be expressible in terms of an operator polynomial; something like (a + D)(D² + D)(ψ³Dψ), where a is a constant and D is ∂/∂x, but I haven't been able to make it work.

I can only suggest putting in trial functions like ψ = sinx and seeing what happens.
Good luck.

 

[Fred Wright]

Why reinvent the wheel when Google is your friend? The Wikipedia page on the KdV equation has c=-1, d=2. Check it out.