KdV Cnoidal Wave Solutions

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Introduction

We will find a solution of the KdV equation for Shallow water waves,

[math] 2H_{0,\tau}+3H_0H_{0,z}+\frac{1}{3}H_{0,zzz}=0[/math]


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](z,\tau)[/math] space

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

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

and substitute into KdV equation then we obtain

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

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


We rearrange and integrate this equation with respect to [math]\xi[/math] to give

[math]\frac{1}{3} \int H_{\xi\xi\xi} d\xi = \int 2V_0 H_{\xi} d\xi - 3\int H H_\xi d\xi[/math]
[math] \Longrightarrow \frac{1}{3}H_{\xi\xi} = 2V_oH - \frac{3}{2}H^2 +D_1[/math]

then multiply [math]H_\xi [/math] to all terms and integrate again


[math]\Longrightarrow \frac{1}{3}H_{\xi\xi} H_\xi = 2V_0HH_\xi - \frac{3H^2}{2}H_\xi + D_1H_\xi[/math]


[math]\Longrightarrow \frac{1}{3} \int H_{\xi\xi}H_\xi d\xi = 2V_o \int H H_\xi d\xi - \frac{3}{2} \int H^2H_\xi d\xi + \int D_1 H_\xi d\xi[/math]


[math]\Longrightarrow \frac{1}{3} \cdot \frac{1}{2}H_\xi^2 = 2V_0 \frac{H^2}{2}-\frac{H^3}{2} +D_1H + D_2 [/math]


[math] \Longrightarrow \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 [math]D_1[/math] and [math]D_2[/math] are constants of integration.

Standardization of KdV equation

We define [math]f(H)= V_oH^2-\frac{1}{2}H^3+D_1H+D2[/math], so [math]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] 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] f(H)= \frac{-1}{2}(H-H_1)(H-H_2)(H-H_3)[/math]


From the equation [math]f(H)=\frac{1}{6}H_\xi^2[/math], we require [math]f(H)\gt0[/math]

We are only interested in solution for [math]H_2 \lt H \lt H_3[/math] and we need [math]H_2 \lt H_3[/math].

and now solve equation in terms of the roots [math]H_i,[/math]

We define [math]X=\frac{H}{H_3}[/math], and obtain

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

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

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

and a further variable Y via

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


[math]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]Y(0)=0.[/math]

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

which is separable.



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

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

Clearly, [math]0 \leq k^2 \leq 1[/math] and [math]l\gt0.[/math]


Solution of the KdV equation

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

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

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


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

for [math]0 \lt k^2 \lt 1 [/math],

or equivalently

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

Now we can write Y with fixed values of [math]x[/math],[math]k[/math] as

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

and hence

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

[math]\mathrm {cn}(x;k)[/math] is another Jacobi elliptic function with [math]\mathrm {cn}^2+\mathrm {sn}^2=1[/math], and waves are called "cnoidal waves".

Using the result [math]cn^2+sn^2=1[/math], our final result can be expressed in the form

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