Perturbation expansion when solving Kdv equation

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Discussion Overview

The discussion revolves around the use of perturbation series in solving the Korteweg-de Vries (KdV) equation, specifically questioning the choice of starting the series with a term of the form eu_1 rather than including a term u_0. Participants explore the implications of this choice and the concept of weak nonlinearity in the context of the KdV equation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions why the perturbation series for the KdV equation is assumed to start with eu_1 instead of u_0 + eu_1.
  • Another participant suggests that including u_0 would lead to it satisfying the original KdV equation, which may not be helpful.
  • A further reply indicates that starting with eu_1 is due to the realization that the O(1) term must satisfy the original equation, implying a lack of physical justification for including u_0.
  • One participant explains that the assumption of a perturbative expansion implies that u(x,t) is not large in amplitude, suggesting a weakly nonlinear system.
  • There is a discussion about the implications of having u(x,t) as O(1) and how this affects the order of derivatives and terms in the equation.
  • A participant seeks clarification on the term "weakly nonlinear" and questions whether it implies that the nonlinear term uu_x is O(epsilon) and how this relates to other terms in the equation.
  • Another participant inquires about the motivation for using a perturbative expansion, questioning whether it aims to cancel nonlinear effects at leading order.

Areas of Agreement / Disagreement

Participants express differing views on the implications of starting the perturbation series with eu_1 versus including u_0. There is no consensus on the motivations behind the perturbative approach or the definition of weak nonlinearity.

Contextual Notes

Participants highlight the dependence on assumptions regarding the amplitude of u(x,t) and the implications for the terms in the KdV equation. The discussion remains open regarding the definitions and implications of weak nonlinearity.

Who May Find This Useful

This discussion may be of interest to those studying nonlinear partial differential equations, particularly in the context of perturbation methods and the KdV equation.

hanson
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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

Please kindly help.
 
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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?
 
Mute said:
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?

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?
 
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).
 
Mute, thanks for your reply first.
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:
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..?
 

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