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Travelling Wave Solutions of the KdV Equation

The KdV equation has two qualitatively different types of permanent form travelling wave solution. These are referred to as cnoidal waves and solitary waves.

Introduction

Assume we have wave travelling with speed [math]\displaystyle{ V_0 }[/math] without change of form,

[math]\displaystyle{ H(z,\tau)=H(z-V_0\tau) }[/math]

and substitute into KdV equation then we obtain

[math]\displaystyle{ -2V_oH_\xi+3HH_\xi+\frac{1}{3}H_{\xi\xi\xi}=0 }[/math]

where [math]\displaystyle{ \xi=z-V_0\tau }[/math] is the travelling wave coordinate.

We integrate this equation twice with respect to [math]\displaystyle{ \xi }[/math] to give

[math]\displaystyle{ \frac{1}{6}H_\xi^2=V_oH^2-\frac{1}{2}H^3+D_1H+D_2=f(H,V_0,D_1,D_2) }[/math]

where D_1 and D_2 are constants of integration.

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

[math]\displaystyle{ f(H)= \frac{-1}{2}(H-H_1)(H-H_2)(H-H_3) }[/math]

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

[math]\displaystyle{ X_\xi^2=3H_3(X-X_1)(X-X_2)(1-X) }[/math]

where [math]\displaystyle{ X_i=\frac{H_i}{H} }[/math]

crest to be at [math]\displaystyle{ \xi=0 and X(0)=0 }[/math]

and a further variable Y via

[math]\displaystyle{ X=1+(X_2-1)sin^2(Y) }[/math]

[math]\displaystyle{ Y_\xi^2=\frac{3}{4}H_3(1-X_1)\left\{ 1-\frac{(1-X_2)}{(1-X_1)}sin(Y)^2 \right\} ...(1) }[/math]

so [math]\displaystyle{ Y(0)=0. }[/math]

and

[math]\displaystyle{ \frac{dY}{d\xi}=\sqrt{l(1-k^2sin^2(Y))}, }[/math]

which is separable.

In order to get this into a completely standard form we define

[math]\displaystyle{ k^2=\frac{1-X_2}{1-X_1}, l=\frac{3}{4}H_3(1-X_1) ...(2) }[/math]

Clearly, [math]\displaystyle{ 0 \leq k^2 \leq 1 }[/math] and [math]\displaystyle{ l\gt 0. }[/math]

A simple quadrature of equation (1) subject to the condition (2) the gives us

[math]\displaystyle{ \int_{0}^\bar{Y} \frac{dS}{\sqrt{l(1-k^2sin^2(Y))}} = \int_{0}^\bar{Y} dS }[/math]

Jacobi elliptic function [math]\displaystyle{ y= sn(x,k) }[/math] can be written in the form

[math]\displaystyle{ x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}} , }[/math]

for [math]\displaystyle{ 0 \lt k^2 \lt 1 }[/math]

,or equivalently

[math]\displaystyle{ x=\int_{0}^{sin^{-1}y} \frac{dS}{\sqrt{1-k^2sin^2(s)}} }[/math]

with fixed values of x,k as