# Perturbation expansion when solving Kdv equation

1. Apr 2, 2007

### hanson

Hi all.
I have actually asked this question in my another thread but seems not many people notice that.
I am going to repeat the question here and hope to get some replies.

The question is about assuming a perturbation series for the KdV equation.

u_t + 6uu_x + u_xxx = 0

Why one would assume a pertubation series of
u = eu_1 + e^2u_2 + e^3u_3 + ...
but not
u = u_0 + eu_1 + e^2u_2+ e^3u_3+... ?

Can someone explain the subtleties inside this to me?

The same thing is obsreved in this article:
www-personal.engin.umich.edu/~jpboyd/op121_boydchennlskdv.pdf
and a paper.
(Please see the figure)
https://www.physicsforums.com/attachment.php?attachmentid=9655&d=1175451667

2. Apr 2, 2007

### Mute

If your expansion included the term u0, then upon collecting terms in powers of e, you would find that u0 has to satisfy the original KdV equation. That wouldn't help you very much, would it?

3. Apr 2, 2007

### hanson

That means there is no physically sound reasons to start the perturbation series by eu1, but just that we found out that the equation for O(1) is u_0 satisfying the originl Kdv, and hence we start from eu1 rather thn u0?

4. Apr 4, 2007

### Mute

The physically sound reason is that in using a pertubative expansion to begin with you're assuming that u(x,t) is not large in amplitude, i.e., that is it less than O(1). In doing so we're assuming our system is weakly nonlinear.

If u(x,t) is O(1), then in order to make the problem weakly nonlinear, we would require that taking an x derivative reduces the order of the term by some small parameter $\epsilon$ - i.e., u_x is an $O(\epsilon)$ term. But if this is the case, then u_xxxx is an $O(\epsilon^4)$ term, which might make the problem less interesting since this term wouldn't contribute much to the dynamics.

Another way for the problem to be weakly nonlinear is to have u(x,t) $O(\epsilon)$. The terms u_1, u_2, etc., are O(1), but their contribution is made small by the powers of the small parameter epsilon multiplying them. Putting this expansion into the equation above and collecting on powers of $\epsilon$, you get (keeping only $O(\epsilon^2)$ and higher terms):

$u_{1t} + u_{1xxxx} = 0$
$u_{2t} + 6uu_{1x} + u_{2xxxx} = 0$

In this case, u_1 satifies the linearized KdV equation, which you can solve to get u_1, and plug into the equation for u_2 and solve (assuming you can find a nice solution).

5. Apr 6, 2007

### hanson

But I am such a layman that I can't grasp the very fundamental things well.
I am wondering if you can explain me at the first place what do you mean by a "weakly nonlinear problem"?
In an equation like
u_t + 6uu_x + u_xxx =0
This is a nonlinear equation because of the term uu_x, right?
By "weakly nonlinear", do you mean uu_x is O(epsilon)? how about u_t and u_xxx? Shall they all be O(spsilon) as well?

If u is O(e), the uu_x is O(^2), and u_t and u_xxx are O(e) as well, does weakly nonlinear mean the nonlinear term is an order less than other terms? So in this case, the diserpsive term u_xxx is NOT weak??

Last edited: Apr 6, 2007
6. Apr 6, 2007

### hanson

and what is the motivation behind using such a perturbative expansive to solve this? Is that we want to cancel the nonlinear effect at the leading order term or what..?