Korteweg - de Vries equation

Korteweg - de Vries equation

The

Korteweg - de Vries equation

is a

partial differential
equation

defined as

u

t

+

6

⁢

u

⁢

u

x

+

u

x

⁢

x

⁢

x

=

0

(1)

where

u

=

u

⁢

(

x

,

t

)

and the subscripts indicate

derivatives

. This
equation arises in hydrodynamics and was originally proposed to
model waves in a canal. In addition to its practical applications,
this equation is quite interesting as an object of mathematical
study. It exhibits interesting

soliton

solutions, has a large
algebra of conserved quantities, and can be solved using methods
of inverse scattering.

1

Travelling Wave Solution

It is easy to exhibit a solution which describes a traveling wave.
To find this solution, one substitutes the ansatz

u

⁢

(

x

,

t

)

=

f

⁢

(

x

-

c

⁢

t

)

1

1

The most general form of a solution to the two-dimensional

wave equation

is

u

⁢

(

x

,

t

)

=

f

⁢

(

x

-

c

⁢

t

)

+

f

⁢

(

x

+

c

⁢

t

)

which describes waves
propagating in both directions with velocity

c

. Here, we will
look for a solution which only describes a wave propagatng in one
direction.

into the equation to obtain the following equation for

f

:

-

c

⁢

f

′

+

6

⁢

f

⁢

f

′

+

f

′′′

=

0

This equation can be written as

(

-

c

⁢

f

+

3

⁢

f

2

+

f

′′

)

′

=

0

and, hence,

-

c

⁢

f

+

3

⁢

f

2

+

f

′′

=

k

for some constant

k

. Multiplying by

f

′

, we can repeat the same
trick:

-

c

⁢

f

⁢

f

′

+

3

⁢

f

2

⁢

f

′

+

f

′

⁢

f

′′

-

k

⁢

f

′

=

(

-

c

⁢

f

2

/

2

+

f

3

+

(

f

′

)

2

-

k

⁢

f

)

′

=

0

hence

-

c

⁢

f

2

/

2

+

f

3

+

(

f

′

)

2

-

k

⁢

f

=

h

or

f

′

=

k

⁢

f

+

c

⁢

f

2

/

2

-

f

3

This can be solved implicitly by an

integral

:

y

=

∫

f

⁢

(

y

)

k

⁢

f

+

c

⁢

f

2

/

2

-

f

3

⁢

𝑑

y

+

C

Since this is an

elliptic integral

, the result is an

elliptic function

.

The solution obtained above is know as a

solitary wave

, or

soliton

.
This term “solitary wave” refers to the fact that this solution describes
a single wave pulse traveling with velocity

c

. Note that the amplitude of
the pulse is determined by its velocity, unlike in the case of linear wave
equations where the velocity of propagation does not depend upon the amplitude.
There are also solutions which describe more than one solitary wave. In
particular, there are solutions in which two of these waves collide and then
re-emerge from the collision.

2

Conserved Currents

It is possible to exhibit integrals whose value is conserved under the
evolution. To construct these quantities, we begin with

vector fields

.
Suppose that

u

is a thrice

differentiable function

of

x

and

t

and
consider the following vector fields:

v

(

1

)

=

(

u

,

3

⁢

u

2

+

u

x

⁢

x

)

v

(

2

)

=

(

u

2

,

4

⁢

u

3

-

u

x

2

+

2

⁢

u

⁢

u

x

⁢

x

)

Computing their

divergences

and rearranging the result,

∂

⁡

v

t

(

1

)

∂

⁡

t

+

∂

⁡

v

t

(

1

)

∂

⁡

x

=

u

t

+

6

⁢

u

⁢

u

x

+

u

x

⁢

x

⁢

x

∂

⁡

v

t

(

2

)

∂

⁡

t

+

∂

⁡

v

t

(

2

)

∂

⁡

x

=

2

⁢

u

⁢

(

u

t

+

6

⁢

u

⁢

u

x

+

u

x

⁢

x

⁢

x

)

.

If

u

happens to satisfy our

differential equation

, then the
divergence of these vector fields will go zero.

To obtain the conserved quantities, we will integrate them over a rectangle in
the

t

,

x

plane with sides parallel to the

x

and

t

axes use Green’s theorem .

∫

t

1

t

2

𝑑

t

⁢

∫

x

1

x

2

𝑑

x

(

u

t

⁢

(

t

,

x

)

+

6

⁢

u

⁢

(

t

,

x

)

⁢

u

x

⁢

(

t

,

x

)

+

u

x

⁢

x

⁢

x

⁢

(

t

,

x

)

)

=

∫

x

1

x

2

𝑑

x

⁢

(

3

⁢

u

2

⁢

(

x

,

t

2

)

+

u

x

⁢

x

⁢

(

x

,

t

2

)

)

+

∫

t

2

t

1

𝑑

t

⁢

u

⁢

(

t

,

x

2

)

+

∫

x

1

x

2

𝑑

x

⁢

(

3

⁢

u

2

⁢

(

x

,

t

1

)

+

u

x

⁢

x

⁢

(

x

,

t

1

)

)

+

∫

t

1

t

2

𝑑

t

⁢

u

⁢

(

t

,

x

1

)

=

3

⁢

∫

x

1

x

2

𝑑

x

⁢

u

2

⁢

(

x

,

t

2

)

-

3

⁢

∫

x

1

x

2

𝑑

x

⁢

u

2

⁢

(

x

,

t

1

)

+

∫

t

1

t

2

𝑑

t

⁢

u

⁢

(

t

,

x

1

)

-

∫

t

1

t

2

𝑑

t

⁢

u

⁢

(

t

,

x

2

)

+

u

x

⁢

(

x

2

,

t

2

)

-

u

x

⁢

(

x

1

,

t

2

)

-

u

x

⁢

(

x

2

,

t

1

)

+

u

x

⁢

(

x

1

,

t

1

)

∫

t

1

t

2

𝑑

t

⁢

∫

x

1

x

2

𝑑

x

2

⁢

u

⁢

(

u

t

⁢

(

t

,

x

)

+

6

⁢

u

⁢

(

t

,

x

)

⁢

u

x

⁢

(

t

,

x

)

+

u

x

⁢

x

⁢

x

⁢

(

t

,

x

)

)

=

∫

x

1

x

2

𝑑

x

⁢

(

4

⁢

u

3

⁢

(

x

,

t

2

)

-

u

x

2

⁢

(

x

,

t

2

)

+

2

⁢

u

⁢

(

x

,

t

2

)

⁢

u

x

⁢

x

⁢

(

x

,

t

2

)

)

+

∫

t

2

t

1

𝑑

t

⁢

u

2

⁢

(

t

,

x

2

)

+

∫

x

2

x

2

𝑑

x

⁢

(

4

⁢

u

3

⁢

(

x

,

t

1

)

-

u

x

2

⁢

(

x

,

t

1

)

+

2

⁢

u

⁢

(

x

,

t

1

)

⁢

u

x

⁢

x

⁢

(

x

,

t

1

)

)

+

∫

t

1

t

2

𝑑

t

⁢

u

2

⁢

(

t

,

x

1

)

=

∫

x

1

x

2

𝑑

x

⁢

(

4

⁢

u

3

⁢

(

x

,

t

2

)

-

3

⁢

u

x

2

⁢

(

x

,

t

2

)

)

-

∫

x

1

x

2

𝑑

x

⁢

(

4

⁢

u

3

⁢

(

x

,

t

1

)

-

3

⁢

u

x

2

⁢

(

x

,

t

1

)

)

+

∫

t

1

t

2

𝑑

t

⁢

u

2

⁢

(

t

,

x

1

)

-

∫

t

1

t

2

𝑑

t

⁢

u

2

⁢

(

t

,

x

2

)

+

u

⁢

(

x

2

,

t

2

)

⁢

u

x

⁢

(

x

2

,

t

2

)

-

u

⁢

(

x

1

,

t

2

)

⁢

u

x

⁢

(

x

1

,

t

2

)

-

u

⁢

(

x

2

,

t

1

)

⁢

u

x

⁢

(

x

2

,

t

1

)

+

u

⁢

(

x

1

,

t

1

)

⁢

u

x

⁢

(

x

1

,

t

1

)

We consider now several popular

boundary conditions

for our equation:

•

Periodic

Suppose that

u

is periodic in

x

with period

p

and
satisfies the Kortwieg - de Vries equation. Then we integrals over

t

cancel against each other as do the endpoint terms, leaving us with

∫

x

1

x

2

𝑑

x

⁢

u

2

⁢

(

x

,

t

1

)

=

∫

x

1

x

2

𝑑

x

⁢

u

2

⁢

(

x

,

t

2

)

.

•

Whole line

Suppose that

u

satisfies the Kortwieg - de Vries equation
and that

u

and its derivatives tend towards zero as

x

→

±

∞

. Then,
taking the limit as

x

1

→

-

∞

and

x

2

→

+

∞

, we conclude that

∫

-

∞

+

∞

𝑑

x

⁢

u

2

⁢

(

x

,

t

1

)

=

∫

-

∞

+

∞

𝑑

x

⁢

u

2

⁢

(

x

,

t

2

)

.

•

Finite

interval

We may also consider our differential equation on
a finite interval and impose suitable boundary conditions at the endpoints.
In this case, the integrals with respect to

t

do not cancel or automatically
vanish as a consequence of the boundary conditions, so they will not, in general,
give conserved quantities. Nevertheless, it is at least still possible to compute
their value solely from the boundary data without having to know how the solution
behaves in the interior.

Starting with the two vector fields given above, it is possible to generate
more such vector fields and more conserved quantities. Since the Lie bracket
of two vector fields with zero divergence also has zero divergence, we may

Title

Korteweg - de Vries equation

Canonical name

KortewegDeVriesEquation

Date of creation

2013-03-22 13:43:20

Last modified on

2013-03-22 13:43:20

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

19

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 35Q53

Synonym

KdV equation

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