Difference between revisions of "KdV Cnoidal Wave Solutions"
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-2V_oH_\xi+3HH_\xi+\frac{1}{3}H_{\xi\xi\xi}=0</math></center> | -2V_oH_\xi+3HH_\xi+\frac{1}{3}H_{\xi\xi\xi}=0</math></center> | ||
where <math>\xi=z-V_0\tau </math> is the travelling wave coordinate. | where <math>\xi=z-V_0\tau </math> is the travelling wave coordinate. | ||
+ | |||
+ | First we integrate with respect to <math>/xi</math>, | ||
+ | |||
'''Explain how this integration works (i.e. from the assignment)''' | '''Explain how this integration works (i.e. from the assignment)''' |
Revision as of 03:43, 17 October 2008
Introduction
We will find a solution of the KdV equation for Shallow water waves,
The KdV equation has two qualitatively different types of permanent form travelling wave solution.
These are referred to as cnoidal waves and solitary waves.
KdV equation in [math]\displaystyle{ (z,\tau) }[/math] space
Assume we have wave travelling with speed [math]\displaystyle{ V_0 }[/math] without change of form,
and substitute into KdV equation then we obtain
where [math]\displaystyle{ \xi=z-V_0\tau }[/math] is the travelling wave coordinate.
First we integrate with respect to [math]\displaystyle{ /xi }[/math],
Explain how this integration works (i.e. from the assignment)
We integrate this equation twice with respect to [math]\displaystyle{ \xi }[/math] to give
where D_1 and D_2 are constants of integration.
Standardization of KdV equation
We define [math]\displaystyle{ f(H)= V_oH^2-\frac{1}{2}H^3+D_1H+D2 }[/math], so [math]\displaystyle{ f(H)=\frac{1}{6}H_\xi^2 }[/math]
It turns out that we require 3 real roots to obtain periodic solutions. Let roots be [math]\displaystyle{ H_1 \leq H_2 \leq H_3 }[/math].
We can imagine the graph of cubic function which has 3 real roots and we can now write a function
From the equation [math]\displaystyle{ f(H)=\frac{1}{6}H_\xi^2 }[/math], we require [math]\displaystyle{ f(H)\gt 0. }[/math]
We are only interested in solution for [math]\displaystyle{ H_2 \lt H \lt H_3 }[/math] and we need [math]\displaystyle{ H_2 \lt H_3 }[/math].
and now solve equation in terms of the roots [math]\displaystyle{ H_i, }[/math]
We define [math]\displaystyle{ X=\frac{H}{H_3} }[/math], and obtain
where [math]\displaystyle{ X_i=\frac{H_i}{H} }[/math]
crest to be at [math]\displaystyle{ \xi=0 }[/math] and [math]\displaystyle{ X(0)=0 }[/math]
and a further variable Y via the relation
[math]\displaystyle{ Y(0)=0. }[/math]
and
which is separable.
In order to get this into a completely standard form we define
Clearly, [math]\displaystyle{ 0 \leq k^2 \leq 1 }[/math] and [math]\displaystyle{ l\gt 0. }[/math]
Solution of the KdV equation
Use [math]\displaystyle{ sin }[/math] insead of [math]\displaystyle{ \sin }[/math] and use [math]\displaystyle{ {\rm sn} }[/math]
A simple quadrature of equation (1) subject to the condition (2) the gives us
Jacobi elliptic function [math]\displaystyle{ y= sn(x,k) }[/math] can be written in the form
or equivalently
Now we can write Y with fixed values of x,k as
and hence
[math]\displaystyle{ cn(x;k) }[/math] is another Jacobi elliptic function with [math]\displaystyle{ cn^2+sn^2=1 }[/math], and waves are called "cnoidal waves".
Using the result [math]\displaystyle{ cn^2+sn^2=1 }[/math], our final result can be expressed in the form