Difference between revisions of "User talk:Wheo001"

From WikiWaves
Jump to navigationJump to search
 
(18 intermediate revisions by the same user not shown)
Line 16: Line 16:
  
  
We integrate this equation twice with respect to <math>\xi</math> to give
+
We rearrange and integrate this equation with respect to <math>\xi</math> to give
 +
<center><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></center>
  
<center><math> \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></center> where D_1 and D_2 are constants of integration.
+
<center><math> \Longrightarrow
 +
\frac{1}{3}H_{\xi\xi} = 2V_oH - \frac{3}{2}H^2 +D_1</math></center>
 +
 
 +
then multiply <math>H_\xi </math> to all terms and integrate again
 +
 
 +
 
 +
 
 +
<center><math>\Longrightarrow
 +
\frac{1}{3}H_{\xi\xi} H_\xi = 2V_0HH_\xi - \frac{3H^2}{2}H_\xi + D_1H_\xi</math></center>
 +
 
 +
 
 +
 
 +
<center><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></center>
 +
 
 +
 
 +
<center><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></center>
 +
 
 +
 
 +
 
 +
<center><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></center> where <math>D_1</math> and <math>D_2</math> are constants of integration.
  
 
==Standardization of KdV equation==
 
==Standardization of KdV equation==
Line 34: Line 57:
  
  
From the equation <math>f(H)=\frac{1}{6}H_\xi^2</math>, we require <math>f(H)>0.</math>
+
From the equation <math>f(H)=\frac{1}{6}H_\xi^2</math>, we require <math>f(H)>0</math>
  
 
We are only interested in solution for <math>H_2 < H < H_3</math> and we need <math>H_2 < H_3</math>.
 
We are only interested in solution for <math>H_2 < H < H_3</math> and we need <math>H_2 < H_3</math>.
Line 45: Line 68:
 
where <math>X_i=\frac{H_i}{H}</math>
 
where <math>X_i=\frac{H_i}{H}</math>
  
crest to be at <math>\xi=0 and X(0)=0</math>
+
crest to be at <math>\xi=0</math> and <math>X(0)=0</math>
  
 
and a further variable Y via
 
and a further variable Y via
  
<center><math> X=1+(X_2-1)sin^2(Y) </math></center>
+
<center><math> X = 1 +(X_2-1) \sin^2 (Y) </math></center>
  
  
<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\}
+
<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>
 
...(1)</math></center>
Line 58: Line 81:
 
so <math>Y(0)=0.</math>
 
so <math>Y(0)=0.</math>
  
and <center><math>\frac{dY}{d\xi}=\sqrt{l(1-k^2sin^2(Y))},</math></center>
+
and <center><math>\frac{dY}{d\xi}=\sqrt{l(1-k^2\sin^2(Y))},</math></center>
 
which is separable.
 
which is separable.
  
Line 70: Line 93:
 
...(2) </math></center>
 
...(2) </math></center>
  
Clearly, <math>0 \leq k^2 \leq 1</math> and  <math>l>0.</math>
+
Clearly, <math>0 \leq k^2 \leq 1</math>  
 +
and  <math>l>0.</math>
  
 
==Solution of the KdV equation==
 
==Solution of the KdV equation==
Line 76: Line 100:
 
A simple quadrature of equation (1) subject to the condition (2) the gives us
 
A simple quadrature of equation (1) subject to the condition (2) the gives us
  
<center><math>\int_{0}^\bar{Y} \frac{dS}{\sqrt{l(1-k^2sin^2(Y))}} = \int_{0}^\bar{Y} dS</math></center>  
+
<center><math>\int_{0}^\bar{Y} \frac{dS}{\sqrt{l(1-k^2\sin^2(Y))}} = \int_{0}^\bar{Y} dS</math></center>  
  
Jacobi elliptic function <math> y= sn(x,k)</math> can be written in the form
+
Jacobi elliptic function <math>y= sn(x,k)</math> can be written in the form
  
  
<center><math> x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}} , for 0 < k^2 < 1</math></center>
+
<center><math> x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}}</math></center>
 +
for <math>0 < k^2 < 1 </math>
  
 
or equivalently
 
or equivalently
  
<center><math> x=\int_{0}^{sin^{-1}y} \frac{dS}{\sqrt{1-k^2sin^2(s)}} </math></center>
+
<center><math> x=\int_{0}^{\sin^{-1}y} \frac{dS}{\sqrt{1-k^2\sin^2(s)}} </math></center>
  
Now we can write Y with fixed values of x,k as
+
Now we can write Y with fixed values of <math>x</math>,<math>k</math> as
  
<center><math> \bar{Y}=sin^{-1}(sn(\sqrt{l}\int_0^{\bar{Y}} d\xi,k)),</math></center>
+
<center><math> \bar{Y}=\sin^{-1}(\mathrm {sn} (\sqrt{l}\int_0^{\bar{Y}} d\xi,k)),</math></center>
  
<center><math> sin(Y)=sn(\sqrt{l}\xi;k),</math></center>
+
<center><math> \sin(Y)=\mathrm {sn}(\sqrt{l}\xi;k),</math></center>
 
and hence  
 
and hence  
<center><math>X=1+(X_2-1)sn^2(\sqrt{l}\xi;k)</math></center>
+
<center><math>X=1+(X_2-1)\mathrm {sn}^2(\sqrt{l}\xi;k)</math></center>
  
<math>cn(x;k)</math> is another Jacobi elliptic function with <math>cn^2+sn^2=1</math>, and waves are called "cnoidal waves".
+
<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
 
Using the result <math>cn^2+sn^2=1</math>, our final result can be expressed in the form
  
<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>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></center>

Latest revision as of 07:19, 17 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.

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

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^2\sin^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}} }[/math]

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

or equivalently

[math]\displaystyle{ 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]\displaystyle{ x }[/math],[math]\displaystyle{ k }[/math] as

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

and hence

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

[math]\displaystyle{ \mathrm {cn}(x;k) }[/math] is another Jacobi elliptic function with [math]\displaystyle{ \mathrm {cn}^2+\mathrm {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)\mathrm {cn}^2\left\{ \left[ \frac{3}{4}(H_3-H_1) \right]^{\frac{1}{2}}\xi;k \right\} }[/math]