Difference between revisions of "Introduction to KdV"

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Line 40: Line 40:
 
If we substitute this expression into the KdV equation we obtain  
 
If we substitute this expression into the KdV equation we obtain  
 
<center><math>
 
<center><math>
-c\frac{df}{d\zeta}+6f\frac{df}{d\zeta}+\frac{d^{3}f}{d\zeta^{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>\zeta</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}{d\zeta^{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.
Line 51: Line 51:
 
V(f) </math> for which the equation is  
 
V(f) </math> for which the equation is  
 
<center><math>
 
<center><math>
\frac{d^{2}f}{d\zeta^{2}}= -\frac{dV}{df}  
+
\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}= -\frac{\mathrm{d}V}{\mathrm{d}f}  
 
</math></center>
 
</math></center>
 
then the potential well is given by  
 
then the potential well is given by  
Line 69: 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}{d\zeta} &=&A_{1}+cf-3f^{2} \\
+
\frac{\mathrm{d}v}{\mathrm{d}\zeta} &=&A_{1}+cf-3f^{2} \\
\frac{df}{d\zeta} &=&v
+
\frac{\mathrm{d}f}{\mathrm{d}\zeta} &=&v
 
\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(
Line 84: Line 84:
 
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}x^{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  
Line 97: Line 97:
 
This is a separable equation and the only challenge is to integrate  
 
This is a separable equation and the only challenge is to integrate  
 
<center><math>
 
<center><math>
\int \frac{1}{\sqrt{-V(f)}} df.
+
\int \frac{1}{\sqrt{-V(f)}} \mathrm{d}f.
 
</math></center>
 
</math></center>
  
Line 109: Line 109:
 
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 d\zeta  
+
\int \frac{\mathrm{d}f}{f\sqrt{c-2f}}=\int \mathrm{d}\zeta  
 
</math></center>
 
</math></center>
 
We then substitute  
 
We then substitute  
Line 121: Line 121:
 
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{c}}\int \frac{\sinh \left(
+
\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{sech}^{2}\left( s\right) \tanh \left( f\right) \cosh ^{3}\left(
f\right) }ds \\
+
f\right) }\mathrm{d}s \\
 
&=&-\frac{2}{\sqrt{c}}s
 
&=&-\frac{2}{\sqrt{c}}s
 
\end{matrix}</math></center>
 
\end{matrix}</math></center>

Revision as of 10:34, 20 August 2010

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 think about this equation as Newton's second law in a potential well [math]\displaystyle{ V(f) }[/math] for which 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}v}{\mathrm{d}\zeta} &=&A_{1}+cf-3f^{2} \\ \frac{\mathrm{d}f}{\mathrm{d}\zeta} &=&v \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). 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.


We can also integrate the equation

[math]\displaystyle{ -cf+3f^{2}+\frac{\mathrm{d}^{2}f}{\mathrm{d}x^{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{-V(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{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 cnodal 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]. 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.