Difference between revisions of "Introduction to KdV"

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The KdV (Korteweg-De Vries) equation is one of the most important non-linear
 
The KdV (Korteweg-De Vries) equation is one of the most important non-linear
 
pde's. It was originally derived to model shallow water waves with weak
 
pde's. It was originally derived to model shallow water waves with weak
nonlinearities, but it has a wide variety of applications. The KdV equation
+
nonlinearities, but it has a wide variety of applications. The derivation of
 +
the KdV is given in [[KdV Equation Derivation]]. The KdV equation
 
is written as  
 
is written as  
 
<center><math>
 
<center><math>
 
\partial _{t}u+6u\partial _{x}u+\partial _{x}^{3}u=0.  
 
\partial _{t}u+6u\partial _{x}u+\partial _{x}^{3}u=0.  
 
</math></center>
 
</math></center>
 +
 +
More information about it can be found at [http://en.wikipedia.org/wiki/Korteweg%E2%80%93de_Vries_equation Korteweg de Vries equation]
  
 
==Travelling Wave Solution==
 
==Travelling Wave Solution==
Line 20: Line 23:
 
The KdV equation posesses travelling wave solutions. One particular
 
The KdV equation posesses travelling wave solutions. One particular
 
travelling wave solution is called a soltion and it was discovered
 
travelling wave solution is called a soltion and it was discovered
experimentally by John Scott Russell in 1834 and was not understoon
+
experimentally by [http://en.wikipedia.org/wiki/John_Scott_Russell John Scott Russell]
theoretically until the work of Korteweg and De Vries in 1895.
+
in 1834. However, it was not understood
 +
theoretically until the work of [http://en.wikipedia.org/wiki/Diederik_Korteweg Korteweg] and  
 +
[http://en.wikipedia.org/wiki/Gustav_de_Vries de Vries] in 1895.
  
 
We begin with the assumption that the wave travels with contant form, i.e.
 
We begin with the assumption that the wave travels with contant form, i.e.
Line 28: Line 33:
 
u\left( x,t\right) =f\left( x-ct\right)  
 
u\left( x,t\right) =f\left( x-ct\right)  
 
</math></center>
 
</math></center>
Note that in this equation the parameter <math>c</math> is an unknown as in the
+
Note that in this equation the parameter <math>c</math> is an unknown as is the
function <math>f.</math> Only very special values of <math>c</math> and <math>f</math> will give travelling
+
function <math>f.</math>  
waves. If we substitute this expression into the KdV equation we obtain  
+
Only very special values of <math>c</math> and <math>f</math> will give travelling
 +
waves.  
 +
We introduce the coordinate <math>\zeta = x - ct</math>.
 +
If we substitute this expression into the KdV equation we obtain  
 
<center><math>
 
<center><math>
-c\frac{df}{dx}+6f\frac{df}{dx}+\frac{d^{3}f}{dx^{3}}=0  
+
-c\frac{\mathrm{d}f}{\mathrm{d}\zeta}+6f\frac{\mathrm{d}f}{\mathrm{d}\zeta}+\frac{\mathrm{d}^{3}f}{\mathrm{d}\zeta^{3}}=0  
 
</math></center>
 
</math></center>
We can integrate this with respect to <math>x</math> to obtain  
+
We can integrate this with respect to <math>\zeta</math> to obtain  
 
<center><math>
 
<center><math>
-cf+3f^{2}+\frac{d^{2}f}{dx^{2}}=A_{1}  
+
-cf+3f^{2}+\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}=A_{1}  
 
</math></center>
 
</math></center>
 
where <math>A</math> is a constant of integration.
 
where <math>A</math> is a constant of integration.
  
If think about this equation as Newton's second law in a potential well <math>
+
If we think about this equation as Newton's second law in a potential well <math>
 
V(f) </math> then the equation is  
 
V(f) </math> then the equation is  
 
<center><math>
 
<center><math>
\frac{d^{2}f}{dx^{2}}=\frac{dV}{df}
+
\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}= -\frac{\mathrm{d}V}{\mathrm{d}f}  
</math></center>
 
We can write this equation as
 
<center><math>
 
\frac{d^{2}f}{dx^{2}}=A_{1}+cf-3f^{2}  
 
 
</math></center>
 
</math></center>
The equation for Newton's second law in a potential well <math>V(f)</math> is given by
+
then the potential well is given by  
 
<center><math>
 
<center><math>
\frac{d^{2}f}{dx^{2}}=-\frac{dV}{df}
+
V\left( f\right) =-A_{0}-A_{1}f-c\frac{f^{2}}{2}+f^{3}  
</math></center>
 
Therefore the potential well is given by
 
<center><math>
 
V\left( f\right) =-A_{0}-A_{1}f-\frac{2}f^{2}+f^{3}  
 
 
</math></center>
 
</math></center>
 
Therefore our equation for <math>f</math> may be thought of as the motion of a particle
 
Therefore our equation for <math>f</math> may be thought of as the motion of a particle
 
in a cubic well.
 
in a cubic well.
  
==Solitary Wave Solution==
+
The constant <math>A_0</math> has no effect on our solution so we can set it to be zero.
 
+
We can choose the constant <math>A_{1}=0</math> and then we have a
We can choose the constants so that <math>A_{0}=A_{1}=0</math> and then we have a
 
 
maximum at <math>f=0</math>. There is a solution which rolls from this at <math>t=-\infty </math>
 
maximum at <math>f=0</math>. There is a solution which rolls from this at <math>t=-\infty </math>
 
and then runs up the other side and finally returns to the maximum at <math>
 
and then runs up the other side and finally returns to the maximum at <math>
Line 70: Line 69:
 
f^{^{\prime }}=v.</math> This gives us  
 
f^{^{\prime }}=v.</math> This gives us  
 
<center><math>\begin{matrix}
 
<center><math>\begin{matrix}
\frac{dv}{dx} &=&A_{1}+cf-3f^{2} \\
+
\frac{\mathrm{d}f}{\mathrm{d}\zeta} &=&v \\
\frac{df}{dx} &=&v
+
\frac{\mathrm{d}v}{\mathrm{d}\zeta} &=&A_{1}+cf-3f^{2}
 
\end{matrix}</math></center>
 
\end{matrix}</math></center>
 
If we chose <math>A_{1}=0</math> then we obtain two equilibria at <math>(f,v)=\left(
 
If we chose <math>A_{1}=0</math> then we obtain two equilibria at <math>(f,v)=\left(
0,0\right) <math> and </math>(3/c,0).</math> If we analysis these equilibria we find the
+
0,0\right) </math> and <math>(3/c,0).</math> If we analysis these equilibria we find the
first is a saddle and the second is difficult to classify. There is a
+
first is a saddle and the second is a nonlinear center (it is neither repelling nor
homoclinic connection which connects the equilibrium point at the origin.
+
attracting).  
 +
The Jacobian matrix for the saddle point is
 +
<center><math>
 +
J_{\left( 0,0\right) }=\left(
 +
\begin{array}{cc}
 +
0 & 1 \\
 +
c & 0
 +
\end{array}
 +
\right)
 +
</math></center>
 +
which has eigenvalues at <math>\pm \sqrt{c}</math> and the incident directions are
 +
<center><math>
 +
\left(
 +
\begin{array}{c}
 +
1\\
 +
\sqrt{c}
 +
\end{array}
 +
\right) \leftrightarrow \sqrt{c},\left(
 +
\begin{array}{c}
 +
-1 \\
 +
\sqrt{c}
 +
\end{array}
 +
\right) \leftrightarrow -\sqrt{c}
 +
</math></center>
 +
There is a
 +
homoclinic connection which connects the equilibrium point at the origin. This holoclinic
 +
connection represents the solitary wave.  Within this homoclinic connection
 +
lie periodic orbits which represent the cnoidal waves.
  
==Formula for the solitary and cnoidal wave.==
+
{| class="wikitable"
 +
|-
 +
! Phase portrait
 +
! Solitary Wave
 +
! Cnoidal Wave
 +
|-
 +
| [[Image:Kdv phase portrait.jpg|thumb|350px|Phase portrait]]
 +
| [[Image:Kdv wave solitary2.gif|thumb|350px|Solitary Wave]]
 +
| [[Image:Kdv wave cn.gif|thumb|350px|Cnoidal Wave]]
 +
|}
  
 
We can also integrate the equation  
 
We can also integrate the equation  
 
<center><math>
 
<center><math>
-cf+3f^{2}+\frac{d^{2}f}{dx^{2}}=A_{1}  
+
-cf+3f^{2}+\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}=A_{1}  
 
</math></center>
 
</math></center>
 
by multiplying by <math>f^{\prime }</math>and integrating. This gives us  
 
by multiplying by <math>f^{\prime }</math>and integrating. This gives us  
 
<center><math>
 
<center><math>
\frac{\left( f^{^{\prime }}\right) ^{2}}{2}=A_{0}+A_{1}f  
+
\frac{\left( f^{^{\prime }}\right) ^{2}}{2}=A_{0}+A_{1}f + c\frac{
 +
f^{2}}{2}-f^{3} = -V(f)
 
</math></center>
 
</math></center>
We write this equation as
+
It is no coincidence that the right hand side is the potential energy, because this
 +
is nothing more that the equation for conservation of energy (or the first
 +
integral of the Lagrangian system) which does not depend on <math>\zeta</math>.
 +
 
 +
This is a separable equation and the only challenge is to integrate
 
<center><math>
 
<center><math>
\frac{\left( f^{^{\prime }}\right) ^{2}}{2}=A_{0}+A_{1}f+\frac{2}
+
\int \frac{1}{\sqrt{-2V(f)}} \mathrm{d}f.
f^{2}-f^{3}=-V(f)
 
 
</math></center>
 
</math></center>
This is nothing more than the equation for conservation of energy for our
+
 
moving particle. We know that the solitary wave solution is found when <math>
+
==Formula for the solitary wave==
 +
 
 +
We know that the solitary wave solution is found when <math>
 
A_{0}=A_{1}=0.</math> This gives us
 
A_{0}=A_{1}=0.</math> This gives us
 
<center><math>
 
<center><math>
Line 101: Line 142:
 
This can be solved by separation of variables to give  
 
This can be solved by separation of variables to give  
 
<center><math>
 
<center><math>
\int \frac{df}{f\sqrt{c-2f}}=\int dx
+
\int \frac{\mathrm{d}f}{f\sqrt{c-2f}}=\int \mathrm{d}\zeta
 
</math></center>
 
</math></center>
 
We then substitute  
 
We then substitute  
 
<center><math>
 
<center><math>
f=\frac{1}{2}c\sec h^{2}\left( s\right)  
+
f=\frac{1}{2}c\,\mathrm{sech}^{2}\left( s\right)  
 
</math></center>
 
</math></center>
 
and note that  
 
and note that  
 
<center><math>
 
<center><math>
c-2f=c\left( 1-\sec h^{2}\left( s\right) \right) =c\tanh ^{2}\left( f\right)  
+
c-2f=c\left( 1-\mathrm{sech}^{2}\left( s\right) \right) =c\tanh ^{2}\left( f\right)  
 
</math></center>
 
</math></center>
 
and that  
 
and that  
 
<center><math>
 
<center><math>
\frac{df}{ds}=-c\frac{\sinh \left( f\right) }{\cosh ^{3}\left( f\right) }  
+
\frac{\mathrm{d}f}{\mathrm{d}s}=-c\frac{\sinh \left( f\right) }{\cosh ^{3}\left( f\right) }  
 
</math></center>
 
</math></center>
 
This means that  
 
This means that  
 
<center><math>\begin{matrix}
 
<center><math>\begin{matrix}
\int \frac{df}{f\sqrt{c-2f}} &=&-\frac{2}{\sqrt}\int \frac{\sinh \left(
+
\int \frac{\mathrm{d}f}{f\sqrt{c-2f}} &=&-\frac{2}{\sqrt{c}}\int \frac{\sinh \left(
f\right) }{\sec h^{2}\left( s\right) \tanh \left( f\right) \cosh ^{3}\left(
+
f\right) }{\mathrm{sech}^{2}\left( s\right) \tanh \left( f\right) \cosh ^{3}\left(
f\right) }ds \\
+
f\right) }\mathrm{d}s \\
&=&-\frac{2}{\sqrt}s
+
&=&-\frac{2}{\sqrt{c}}s
 
\end{matrix}</math></center>
 
\end{matrix}</math></center>
 
This gives us
 
This gives us
 
<center><math>
 
<center><math>
-\frac{2}{\sqrt}s=x+a  
+
-\frac{2}{\sqrt{c}}s=\zeta+a  
 
</math></center>
 
</math></center>
 
Therefore  
 
Therefore  
 
<center><math>
 
<center><math>
f=\frac{1}{2}c\sec h^{2}\left( \frac{\sqrt}{2}\left( x+a\right) \right)  
+
f=\frac{1}{2}c\,\mathrm{sech}^{2}\left( \frac{\sqrt{c}}{2}\left( \zeta+a\right) \right)  
 
</math></center>
 
</math></center>
 
Of course we assumed that <math>x=x-ct</math> so the formula for the solitary wave is
 
Of course we assumed that <math>x=x-ct</math> so the formula for the solitary wave is
 
given by
 
given by
 
 
<center><math>
 
<center><math>
f\left( x-ct\right) =\frac{1}{2}c\sec h^{2}\left[ \frac{\sqrt}{2}\left(
+
f\left( x-ct\right) =\frac{1}{2}c\,\mathrm{sech}^{2}\left[ \frac{\sqrt{c}}{2}\left(
 
x-ct+a\right) \right]  
 
x-ct+a\right) \right]  
 
</math></center>
 
</math></center>
We have a cubic well potential. \ One solution is where the ball rolls from
+
Note that a solution exists for each
the top and exactly returns to the maximum. The second is where the ball
+
<math>c</math>, and that the amplitude is proportional to <math>c.</math> All of this was
rolls back and forth. Formulas (not simple) can be derived for these
+
discovered experimentally by
solutions. The solitary wave is given by
+
[http://en.wikipedia.org/wiki/John_Scott_Russell John Scott Russell].
 +
 
 +
==Formula for the cnoidal wave==
 +
 
 +
If we consider the case when the solution oscillates between two values <math>F_2 < F_3</math>
 +
(which we can assume are also roots of <math>V(f)</math> without loss of generality) then
 +
we can integrate the equation to obtain
 
<center><math>
 
<center><math>
f\left( x-ct\right) =\frac{1}{2}c\sec h^{2}\left[ \frac{\sqrt}{2}\left(
+
f\left( \zeta\right) =F_2+(F_3 - F_2) \mathrm{cn}^{2}\left( \gamma \zeta ;k\right)  
x-ct+a\right) \right]
 
 
</math></center>
 
</math></center>
which describes a right moving soliton. Note that a solution exists for each
+
where <math>cn</math> is a Jacobi Elliptic function and
<math>c</math>, and that the amplidute is proportional to <math>c.</math> All of this was
+
<math>\gamma</math> and <math>k</math> are constants which depend on <math>c</math>.
discovered experimentally by Russel. We also have cnoidal wave solutions,
+
Derivation of this equation is found [[KdV Cnoidal Wave Solutions]].
which are periodic waves, of the form
+
We can write this equation as
 
<center><math>
 
<center><math>
f\left( x-ct\right) =a+bcn^{2}\left( x-ct\right)  
+
f\left( x-ct\right) =a+b \mathrm{cn}^{2}\left( \gamma (x-ct);k\right)  
 
</math></center>
 
</math></center>
where <math>cn</math> is a Jacobi Elliptic function. In the limit the two solution
+
where <math>a=k^2\gamma^2</math> and <math>c = 6b + 4(2k^2 -1)\gamma^2</math>.
agree.
+
These waves are known as [http://en.wikipedia.org/wiki/Cnoidal_wave cnoidal waves].
 +
 
 +
In the limit the two solutions agree. We also obtain a sinusoidal solution in the limit of
 +
small amplitude.
 +
 
 +
== Lecture Videos ==
 +
 
 +
=== Part 1 ===
 +
 
 +
{{#ev:youtube|SzZ-KhvvPio}}
 +
 
 +
=== Part 2 ===
 +
 
 +
{{#ev:youtube|QltlSQQBtrs}}
 +
 
 +
=== Part 3 ===
 +
 
 +
{{#ev:youtube|NJ7h3Z9QtvU}}
 +
 
 +
=== Part 4 ===
 +
 
 +
{{#ev:youtube|NictSlSgRbM}}

Latest revision as of 03:57, 10 August 2023

Nonlinear PDE's Course
Current Topic Introduction to KdV
Next Topic Numerical Solution of the KdV
Previous Topic Nonlinear Shallow Water Waves




The KdV (Korteweg-De Vries) equation is one of the most important non-linear pde's. It was originally derived to model shallow water waves with weak nonlinearities, but it has a wide variety of applications. The derivation of the KdV is given in KdV Equation Derivation. The KdV equation is written as

[math]\displaystyle{ \partial _{t}u+6u\partial _{x}u+\partial _{x}^{3}u=0. }[/math]

More information about it can be found at Korteweg de Vries equation

Travelling Wave Solution

The KdV equation posesses travelling wave solutions. One particular travelling wave solution is called a soltion and it was discovered experimentally by John Scott Russell in 1834. However, it was not understood theoretically until the work of Korteweg and de Vries in 1895.

We begin with the assumption that the wave travels with contant form, i.e. is of the form

[math]\displaystyle{ u\left( x,t\right) =f\left( x-ct\right) }[/math]

Note that in this equation the parameter [math]\displaystyle{ c }[/math] is an unknown as is the function [math]\displaystyle{ f. }[/math] Only very special values of [math]\displaystyle{ c }[/math] and [math]\displaystyle{ f }[/math] will give travelling waves. We introduce the coordinate [math]\displaystyle{ \zeta = x - ct }[/math]. If we substitute this expression into the KdV equation we obtain

[math]\displaystyle{ -c\frac{\mathrm{d}f}{\mathrm{d}\zeta}+6f\frac{\mathrm{d}f}{\mathrm{d}\zeta}+\frac{\mathrm{d}^{3}f}{\mathrm{d}\zeta^{3}}=0 }[/math]

We can integrate this with respect to [math]\displaystyle{ \zeta }[/math] to obtain

[math]\displaystyle{ -cf+3f^{2}+\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}=A_{1} }[/math]

where [math]\displaystyle{ A }[/math] is a constant of integration.

If we think about this equation as Newton's second law in a potential well [math]\displaystyle{ V(f) }[/math] then the equation is

[math]\displaystyle{ \frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}= -\frac{\mathrm{d}V}{\mathrm{d}f} }[/math]

then the potential well is given by

[math]\displaystyle{ V\left( f\right) =-A_{0}-A_{1}f-c\frac{f^{2}}{2}+f^{3} }[/math]

Therefore our equation for [math]\displaystyle{ f }[/math] may be thought of as the motion of a particle in a cubic well.

The constant [math]\displaystyle{ A_0 }[/math] has no effect on our solution so we can set it to be zero. We can choose the constant [math]\displaystyle{ A_{1}=0 }[/math] and then we have a maximum at [math]\displaystyle{ f=0 }[/math]. There is a solution which rolls from this at [math]\displaystyle{ t=-\infty }[/math] and then runs up the other side and finally returns to the maximum at [math]\displaystyle{ t=\infty . }[/math] This corresponds to a solitary wave solution.

We can also think about the equation as a first order system using [math]\displaystyle{ f^{^{\prime }}=v. }[/math] This gives us

[math]\displaystyle{ \begin{matrix} \frac{\mathrm{d}f}{\mathrm{d}\zeta} &=&v \\ \frac{\mathrm{d}v}{\mathrm{d}\zeta} &=&A_{1}+cf-3f^{2} \end{matrix} }[/math]

If we chose [math]\displaystyle{ A_{1}=0 }[/math] then we obtain two equilibria at [math]\displaystyle{ (f,v)=\left( 0,0\right) }[/math] and [math]\displaystyle{ (3/c,0). }[/math] If we analysis these equilibria we find the first is a saddle and the second is a nonlinear center (it is neither repelling nor attracting). The Jacobian matrix for the saddle point is

[math]\displaystyle{ J_{\left( 0,0\right) }=\left( \begin{array}{cc} 0 & 1 \\ c & 0 \end{array} \right) }[/math]

which has eigenvalues at [math]\displaystyle{ \pm \sqrt{c} }[/math] and the incident directions are

[math]\displaystyle{ \left( \begin{array}{c} 1\\ \sqrt{c} \end{array} \right) \leftrightarrow \sqrt{c},\left( \begin{array}{c} -1 \\ \sqrt{c} \end{array} \right) \leftrightarrow -\sqrt{c} }[/math]

There is a homoclinic connection which connects the equilibrium point at the origin. This holoclinic connection represents the solitary wave. Within this homoclinic connection lie periodic orbits which represent the cnoidal waves.

Phase portrait Solitary Wave Cnoidal Wave
Phase portrait
Solitary Wave
Cnoidal Wave

We can also integrate the equation

[math]\displaystyle{ -cf+3f^{2}+\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}=A_{1} }[/math]

by multiplying by [math]\displaystyle{ f^{\prime } }[/math]and integrating. This gives us

[math]\displaystyle{ \frac{\left( f^{^{\prime }}\right) ^{2}}{2}=A_{0}+A_{1}f + c\frac{ f^{2}}{2}-f^{3} = -V(f) }[/math]

It is no coincidence that the right hand side is the potential energy, because this is nothing more that the equation for conservation of energy (or the first integral of the Lagrangian system) which does not depend on [math]\displaystyle{ \zeta }[/math].

This is a separable equation and the only challenge is to integrate

[math]\displaystyle{ \int \frac{1}{\sqrt{-2V(f)}} \mathrm{d}f. }[/math]

Formula for the solitary wave

We know that the solitary wave solution is found when [math]\displaystyle{ A_{0}=A_{1}=0. }[/math] This gives us

[math]\displaystyle{ \left( f^{^{\prime }}\right) ^{2}=f^{2}\left( c-2f\right) }[/math]

This can be solved by separation of variables to give

[math]\displaystyle{ \int \frac{\mathrm{d}f}{f\sqrt{c-2f}}=\int \mathrm{d}\zeta }[/math]

We then substitute

[math]\displaystyle{ f=\frac{1}{2}c\,\mathrm{sech}^{2}\left( s\right) }[/math]

and note that

[math]\displaystyle{ c-2f=c\left( 1-\mathrm{sech}^{2}\left( s\right) \right) =c\tanh ^{2}\left( f\right) }[/math]

and that

[math]\displaystyle{ \frac{\mathrm{d}f}{\mathrm{d}s}=-c\frac{\sinh \left( f\right) }{\cosh ^{3}\left( f\right) } }[/math]

This means that

[math]\displaystyle{ \begin{matrix} \int \frac{\mathrm{d}f}{f\sqrt{c-2f}} &=&-\frac{2}{\sqrt{c}}\int \frac{\sinh \left( f\right) }{\mathrm{sech}^{2}\left( s\right) \tanh \left( f\right) \cosh ^{3}\left( f\right) }\mathrm{d}s \\ &=&-\frac{2}{\sqrt{c}}s \end{matrix} }[/math]

This gives us

[math]\displaystyle{ -\frac{2}{\sqrt{c}}s=\zeta+a }[/math]

Therefore

[math]\displaystyle{ f=\frac{1}{2}c\,\mathrm{sech}^{2}\left( \frac{\sqrt{c}}{2}\left( \zeta+a\right) \right) }[/math]

Of course we assumed that [math]\displaystyle{ x=x-ct }[/math] so the formula for the solitary wave is given by

[math]\displaystyle{ f\left( x-ct\right) =\frac{1}{2}c\,\mathrm{sech}^{2}\left[ \frac{\sqrt{c}}{2}\left( x-ct+a\right) \right] }[/math]

Note that a solution exists for each [math]\displaystyle{ c }[/math], and that the amplitude is proportional to [math]\displaystyle{ c. }[/math] All of this was discovered experimentally by John Scott Russell.

Formula for the cnoidal wave

If we consider the case when the solution oscillates between two values [math]\displaystyle{ F_2 \lt F_3 }[/math] (which we can assume are also roots of [math]\displaystyle{ V(f) }[/math] without loss of generality) then we can integrate the equation to obtain

[math]\displaystyle{ f\left( \zeta\right) =F_2+(F_3 - F_2) \mathrm{cn}^{2}\left( \gamma \zeta ;k\right) }[/math]

where [math]\displaystyle{ cn }[/math] is a Jacobi Elliptic function and [math]\displaystyle{ \gamma }[/math] and [math]\displaystyle{ k }[/math] are constants which depend on [math]\displaystyle{ c }[/math]. Derivation of this equation is found KdV Cnoidal Wave Solutions. We can write this equation as

[math]\displaystyle{ f\left( x-ct\right) =a+b \mathrm{cn}^{2}\left( \gamma (x-ct);k\right) }[/math]

where [math]\displaystyle{ a=k^2\gamma^2 }[/math] and [math]\displaystyle{ c = 6b + 4(2k^2 -1)\gamma^2 }[/math]. These waves are known as cnoidal waves.

In the limit the two solutions agree. We also obtain a sinusoidal solution in the limit of small amplitude.

Lecture Videos

Part 1

Part 2

Part 3

Part 4