KdV Cnoidal Wave Solutions

Travelling Wave Solutions of the KdV Equation

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,

[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.

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

[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]

Solution of the KdV equation

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}} , for 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]

Now we can write Y with fixed values of x,k as

[math]\displaystyle{ \bar{Y}=sin^{-1}(sn(\sqrt{l}\int_0^{\bar{Y}} d\xi,k)), }[/math] [math]\displaystyle{ sin(Y)=sn(\sqrt{l}\xi;k), }[/math]

and hence

[math]\displaystyle{ X=1+(X_2-1)sn^2(\sqrt{l}\xi;k) }[/math]

[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

[math]\displaystyle{ H=H_2+(H_3-H_2)cn^2\left\{ \left[ \frac{3}{4}(H_3-H_1) \right]^{\frac{1}{2}}\xi;k \right\} }[/math]