Difference between revisions of "KdV Cnoidal Wave Solutions"

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==Solution of the KdV equation==
 
==Solution of the KdV equation==
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Use <math>sin</math> insead of <math>\sin</math> and use <math>{\rm sn}</math>
 
  
−
A simple quadrature of equation (1) subject to the condition (2) the gives us
+
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>
  
−
<center><math>\int_{0}^\bar{Y} \frac{dS}{\sqrt{l(1-k^2sin^2(Y))}} = \int_{0}^\bar{Y} dS</math></center>
+
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>.
  
−
Jacobi elliptic function <math> y= sn(x,k)</math> can be written in the form
 
  
 +
We can imagine the graph of cubic function which has 3 real roots and we can now write a function
 +
<center><math> f(H)= \frac{-1}{2}(H-H_1)(H-H_2)(H-H_3)</math></center>
  
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<center><math> x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}} , for 0 < k^2 < 1</math></center>
 
  
−
or equivalently
+
From the equation <math>f(H)=\frac{1}{6}H_\xi^2</math>, we require <math>f(H)>0</math>
  
−
<center><math> x=\int_{0}^{\sin^{-1}y} \frac{dS}{\sqrt{1-k^2\sin^2(s)}} </math></center>
+
We are only interested in solution for <math>H_2 < H < H_3</math> and we need <math>H_2 < H_3</math>.
  
−
Now we can write Y with fixed values of x,k as
+
and now solve equation in terms of the roots <math>H_i,</math>
  
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<center><math> \bar{Y}=sin^{-1}(sn(\sqrt{l}\int_0^{\bar{Y}} d\xi,k)),</math></center>
+
We define <math>X=\frac{H}{H_3}</math>, and obtain
  
−
<center><math> sin(Y)=sn(\sqrt{l}\xi;k),</math></center>
+
<center><math>X_\xi^2=3H_3(X-X_1)(X-X_2)(1-X)</math></center>
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and hence
+
where <math>X_i=\frac{H_i}{H}</math>
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<center><math>X=1+(X_2-1)sn^2(\sqrt{l}\xi;k)</math></center>
 
  
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<math>cn(x;k)</math> is another Jacobi elliptic function with <math>cn^2+sn^2=1</math>, and waves are called "cnoidal waves".
+
crest to be at <math>\xi=0</math> and <math>X(0)=0</math>
  
−
Using the result <math>cn^2+sn^2=1</math>, our final result can be expressed in the form
+
and a further variable Y via
  
−
<center><math>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></center>
+
<center><math> X = 1 +(X_2-1) \sin^2 (Y) </math></center>
  
−
[[Category:789]]
+
 
 +
<center><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></center>
 +
 
 +
so <math>Y(0)=0.</math>
 +
 
 +
and <center><math>\frac{dY}{d\xi}=\sqrt{l(1-k^2\sin^2(Y))},</math></center>
 +
which is separable.
 +
 
 +
 
 +
 
 +
 
 +
In order to get this into a completely standard form we define
 +
 
 +
<center><math>k^2=\frac{1-X_2}{1-X_1}, l=\frac{3}{4}H_3(1-X_1)
 +
 
 +
...(2) </math></center>
 +
 
 +
Clearly, <math>0 \leq k^2 \leq 1</math>
 +
and  <math>l>0.</math>

Revision as of 07:30, 17 October 2008

Introduction

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

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

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

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


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


[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ D_1 }[/math] and [math]\displaystyle{ D_2 }[/math] 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 }[/math] and [math]\displaystyle{ 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^2\sin^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

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 }[/math] and [math]\displaystyle{ 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^2\sin^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]

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