Must solitary wave a product of weakly nonlinear and weakly dispersive?

hanson
Messages
312
Reaction score
0
Hi all!
I know that KdV is a balance of the nonlinearity and the dispersive effect and hence the wave profile propagates permenantly without disperse.
However, why one always mention that it is the balance of WEAKLY nonlinear and WEAKLY dispersive?
Can't it be a balance of STRONGLY nonlinear and STRONLY dispersive?
Will the KdV equation still exist if the shallow water parameter and the nonlinear parameter does not tend to zero but just with the same order of magnitude? Nnamely O(d^2) = O(epsilon), yet, not tends to zero.

Can someone kindly help?
 
Physics news on Phys.org
Perhaps.
And then again, perhaps not.

Remember that the only reason why we split up problems as "weakly" non-liear, or "weakly" linearly dispersive is in order to justify chopping away some terms so that the remaining image becomes clear and comprehensible.

If all effects should be taken into account, then we are back in a morass of terms and possible outcomes.
 
oh...i see..
That's means weakly nonlinear and weakly linearly dispersive region is not the only region that a KdV balance may occur?
 
From what I know, you may indeed have KdV-like equations with "stronger" non-linearities than of the KdV-type.
 
arildno said:
From what I know, you may indeed have KdV-like equations with "stronger" non-linearities than of the KdV-type.

Thank you arildno.
One more question regarding KdV equation.
I am currently reading an article:
www-personal.engin.umich.edu/~jpboyd/op121_boydchennlskdv.pdf
and a paper. (Please see the attached figure)
Both of them assume the perturbation series of the velocity start from O(epsilon), how come?
It is somewhat counter-intuitive to me since we always assume something like
u ~ u0 + eu1+ e^2u2 +... isn't it?
why does this start from eu1 without the constant velocity u0?
Can you please explain?
 

Attachments

  • why.JPG
    why.JPG
    33.9 KB · Views: 791
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top