KdV Cnoidal Wave Solutions

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Introduction

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

$2H_{0,\tau}+3H_0H_{0,z}+\frac{1}{3}H_{0,zzz}=0$

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 $(z,\tau)$ space

Assume we have wave travelling with speed $V_0$ without change of form,

$H(z,\tau)=H(z-V_0\tau)$

and substitute into KdV equation then we obtain

$-2V_oH_\xi+3HH_\xi+\frac{1}{3}H_{\xi\xi\xi}=0$

where $\xi=z-V_0\tau$ is the travelling wave coordinate.

We rearrange and integrate this equation with respect to $\xi$ to give

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

then multiply $H_\xi$ to all terms and integrate again

$\Longrightarrow \frac{1}{3}H_{\xi\xi} H_\xi = 2V_0HH_\xi - \frac{3H^2}{2}H_\xi + D_1H_\xi$

$\Longrightarrow \frac{1}{3} \int H_{\xi\xi}H_\xi d\xi = 2V_o \int H H_\xi d\xi - \frac{3}{2} \int H^2H_\xi d\xi + \int D_1 H_\xi d\xi$

$\Longrightarrow \frac{1}{3} \cdot \frac{1}{2}H_\xi^2 = 2V_0 \frac{H^2}{2}-\frac{H^3}{2} +D_1H + D_2$

$\Longrightarrow \frac{1}{6}H_\xi^2=V_oH^2-\frac{1}{2}H^3+D_1H+D_2=f(H,V_0,D_1,D_2)$ where $D_1$ and $D_2$ are constants of integration.

Standardization of KdV equation

We define $f(H)= V_oH^2-\frac{1}{2}H^3+D_1H+D2$, so $f(H)=\frac{1}{6}H_\xi^2$

It turns out that we require 3 real roots to obtain periodic solutions. Let roots be $H_1 \leq H_2 \leq H_3$.

We can imagine the graph of cubic function which has 3 real roots and we can now write a function

$f(H)= \frac{-1}{2}(H-H_1)(H-H_2)(H-H_3)$

From the equation $f(H)=\frac{1}{6}H_\xi^2$, we require $f(H)>0$

We are only interested in solution for $H_2 < H < H_3$ and we need $H_2 < H_3$.

and now solve equation in terms of the roots $H_i,$

We define $X=\frac{H}{H_3}$, and obtain

$X_\xi^2=3H_3(X-X_1)(X-X_2)(1-X)$

where $X_i=\frac{H_i}{H}$

crest to be at $\xi=0$ and $X(0)=0$

and a further variable Y via

$X = 1 +(X_2-1) \sin^2 (Y)$

$Y_\xi^2=\frac{3}{4}H_3(1-X_1)\left\{ 1-\frac{(1-X_2)}{(1-X_1)}\sin(Y)^2 \right\} ...(1)$

so $Y(0)=0.$

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

which is separable.

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

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

Clearly, $0 \leq k^2 \leq 1$ and $l>0.$

Solution of the KdV equation

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

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

Jacobi elliptic function $y= sn(x,k)$ can be written in the form

$x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}}$

for $0 < k^2 < 1$,

or equivalently

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

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

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

and hence

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

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

Using the result $cn^2+sn^2=1$, our final result can be expressed in the form

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