﻿ Traveling Wave Solutions to the Regularized Long Wave Equation (RLWE)
San José State University

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 The Traveling Wave Solutions to the Regularized Long Wave Equation (RLWE)

The Regularized Long Wave Equation (RLWE) has various forms. The form which will be utilized in the analysis here is:

#### ut + ux + uux - utxx = 0 which is equivalent to ut + ux + (½u²)x - utxx = 0

The Korteweg-de Vries (KdV) equation may be expressed as

#### ut + ux + uux +uxxx = 0 or, equivalently, to show its relationship to the RLWE ut + ux + (½u²)x - uxxx = 0

The KdV equation derives from the analysis of Korteweg and de Vries in 1895 to derive an equation for water waves 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.

The RLW equation was formulated by Peregrine (1966) as an alternative to KdV equation for studying soliton phenomenon. It was proposed because it would not have the same limitations for the size of the time step in numerical solution that the KdV has.

## Traveling Wave Solutions

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

#### (1 −v)uz + ½(u²)z + vuzzz = 0

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

#### (1 −v)u + ½u² + vuzz = c0

where c0 is an arbitrary constant.

The above equation may be multiplied by uz to give

#### (1 −v)uuz + ½u²uz + vuzzuz = c0uzwhich is the same as (1−v)½(u²)z + 1/6(u³)z + v(½uz²)z = c0uz

This obviously can be integrated with respect to z to give

#### (1−v)½u² + 1/6u³ + ½vuz² = 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:

Therefore

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

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

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

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

#### (1-v) + 4vb² = 0 and a − 12vb² = 0

These conditions imply that

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

There is a forbidden zone for a. Since 4b²=a/(a+3) must be positive it means a must be less than -3 or greater than 0. This means that v is either greater that 1 or less than 0.

(To be continued.)