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The Traveling Wave Solutions
to the Korteweg de Vries Equation
(KdV)

The Korteweg de Vries Equation (KdV) has various forms. It may be expressed as


ut + ux + uux + uxxx = 0
 

The KdV derives from the analysis of Korteweg and de Vries in 1895 to derive an equation for water waters that would explain the existence of a smoothly humped wave observed in nature. When the KdV equation was used in numerical simulations in the 1950's the investigators found that the wave solutions persisted after interactions. These wave solutions were called solitons.

Traveling Wave Solutions

Some aspects of the solutions to the KdV equations may be derived from analysis. A traveling wave solution is of the form u(x-vt-x0). Letting z=x-vt-x0 the KdV equation becomes:


(1 −v)uz + ½(u²)z + uzzz = 0
 

This may be immediately integrated with respect to z to give:


(1 −v)u + ½u² + uzz = c0
 

where c0 is an arbitrary constant.

The above equation may be multiplied by uz to give


(1 −v)uuz + ½u²uz + uzzuz = c0uz
which is the same as
(1−v)½(u²)z + 1/6(u³)z + (½uz²)z = c0uz
 

This obviously can be integrated with respect to z to give


(1−v)½u² + 1/6u³ + ½uz² = c0u + c1
 

In principle this equation could be solved for uz and the result integrated. As a practical matter this would be too cumbersome. Instead let us check to see if U(z) = a·sech²(bz) is a solution to


(1 −v)uz + ½(u²)z + vuzzz = 0
or, equivalently
(1 −v)uz + uuz + vuzzz = 0
 

The terms of this equation for U(z) can be evaluated


Uz = 2a·sech(bz)(dsech(bz)/dz)b = -2ab·sech²(bz)tanh(bz)
and hence
UUz = -2a²b·sech4(bz)tanh(bz)
 

The second derivative is given by:


Uzz = 4ab²sech²(bz)tanh²(bz) - 2ab²sech4(bz)
but since tanh²(bz)=1−sech²(bz)
Uzz = 4ab²sech²(bz) - 6ab²sech4(bz)
 

Therefore


Uzzz = −8ab³sech²(bz)tanh(bz) + 24ab³sech4(bz)tanh(bz)
 

Substituting these expressions into the KdV equation and dividing by -2ab·sech²(bz)tanh(bz) gives


(1-v)+4vb² + (a − 12vb²)sech²(bz) = 0
 

Thus for the KdV equation to be satisfied for all z it must hold that


(1-v) + 4vb² = 0
and
a − 12b² = 0
 

These conditions imply that


b = ½[a/12]1/2
and
v = 1/(1-4b²) = 1/(1 - a/3)
 

The parameter a represents the amplitude of the wave. Parameter b represents the inverse of the width of the wave. The parameter v is the speed of the wave. A positive value of v indicates movement to the right and a negative value movement to the left. Once any one of the three parameters is specified the other two are determined.

(To be continued.)


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