Difference between revisions of "User talk:Wheo001"

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<center><math>
 
<center><math>
-2V_oH_\xi+3HH_xi+\frac{1}{3}H_\xi\xi\xi=0</math></center>
+
-2V_oH_\xi+3HH_xi+\frac{1}{3}H_\xixixi=0</math></center>
 
where <math>xi=z-V_0\tau </math> is the travelling wave coordinate.
 
where <math>xi=z-V_0\tau </math> is the travelling wave coordinate.
 
We integrate this equation twice with respect to <math> xi</math> to give
 
We integrate this equation twice with respect to <math> xi</math> to give

Revision as of 02:07, 14 October 2008

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 substitue into KdV equation then we obtain

[math]\displaystyle{ -2V_oH_\xi+3HH_xi+\frac{1}{3}H_\xixixi=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+D2=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], i.e [math]\displaystyle{ f(H)=\frac{1}{6}H_\xi^2. It turns out that we require 3 real roots to obatain periodic solutions. Let roots be \lt math\gt 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]