Introduction to KdV

Nonlinear PDE's Course
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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 KdV equation is written as

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

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 and was not understoon 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 in the function [math]\displaystyle{ f. }[/math] Only very special values of [math]\displaystyle{ c }[/math] and [math]\displaystyle{ f }[/math] will give travelling waves. If we substitute this expression into the KdV equation we obtain

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

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

[math]\displaystyle{ -cf+3f^{2}+\frac{d^{2}f}{dx^{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] then the equation is

[math]\displaystyle{ \frac{d^{2}f}{dx^{2}}=\frac{dV}{df} }[/math]

We can write this equation as

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

The equation for Newton's second law in a potential well [math]\displaystyle{ V(f) }[/math] is given by

[math]\displaystyle{ \frac{d^{2}f}{dx^{2}}=-\frac{dV}{df} }[/math]

Therefore the potential well is given by

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

Solitary Wave Solution

We can choose the constants so that [math]\displaystyle{ A_{0}=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{dv}{dx} &=&A_{1}+cf-3f^{2} \\ \frac{df}{dx} &=&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 difficult to classify. There is a homoclinic connection which connects the equilibrium point at the origin.

Formula for the solitary and cnoidal wave.

We can also integrate the equation

[math]\displaystyle{ -cf+3f^{2}+\frac{d^{2}f}{dx^{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 }[/math]

We write this equation as

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

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]\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{df}{f\sqrt{c-2f}}=\int dx }[/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{df}{ds}=-c\frac{\sinh \left( f\right) }{\cosh ^{3}\left( f\right) } }[/math]

This means that

[math]\displaystyle{ \begin{matrix} \int \frac{df}{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) }ds \\ &=&-\frac{2}{\sqrt{c}}s \end{matrix} }[/math]

This gives us

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

Therefore

[math]\displaystyle{ f=\frac{1}{2}c\mathrm{sech}^{2}\left( \frac{\sqrt{c}}{2}\left( x+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]

We have a cubic well potential. One solution is where the ball rolls from the top and exactly returns to the maximum. The second is where the ball rolls back and forth. Formulas (not simple) can be derived for these solutions. 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]

which describes a right moving soliton. Note that a solution exists for each [math]\displaystyle{ c }[/math], and that the amplidute is proportional to [math]\displaystyle{ c. }[/math] All of this was discovered experimentally by Russel. We also have cnoidal wave solutions, which are periodic waves, of the form

[math]\displaystyle{ f\left( x-ct\right) =a+bcn^{2}\left( x-ct\right) }[/math]

where [math]\displaystyle{ cn }[/math] is a Jacobi Elliptic function. In the limit the two solution agree.