example of eliminating higher-order derivatives

example of eliminating higher-order derivatives

To show how a

partial differential equation

involving

partial derivatives

of order higher than the first can be re-expressed as a system involving only

first derivatives

to which the simple form of the

Cauchy-Kowalewski theorem

can be applied, consider the following example:

∂

2

⁡

u

∂

⁡

t

2

=

f

⁢

(

∂

2

⁡

u

∂

⁡

x

2

,

∂

2

⁡

u

∂

⁡

t

⁢

∂

⁡

x

)

First, the second derivative with respect to

t

can be eliminated by introducing a new variable

u

t

and a adding a

differential equation

which sets

u

t

equal to the

derivative

of

u

.

∂

⁡

u

t

∂

⁡

t

=

f

⁢

(

∂

2

⁡

u

∂

⁡

x

2

,

∂

⁡

u

t

∂

⁡

x

)

∂

⁡

u

∂

⁡

t

=

u

t

To eliminate the second derivatives involving

x

, we first introduce variables

u

x

⁢

x

,

u

t

⁢

x

and

u

t

⁢

t

. To obtain an equation with

∂

⁡

u

t

⁢

t

/

∂

⁡

t

on the left-hand side, differentiate the original equation with respect to

t

∂

3

⁡

u

∂

⁡

t

3

=

f

′

1

⁢

(

∂

2

⁡

u

∂

⁡

x

2

,

∂

2

⁡

u

∂

⁡

t

⁢

∂

⁡

x

)

⁢

∂

3

⁡

u

∂

⁡

t

⁢

∂

⁡

x

2

+

f

′

2

⁢

(

∂

2

⁡

u

∂

⁡

x

2

,

∂

2

⁡

u

∂

⁡

t

⁢

∂

⁡

x

)

⁢

∂

3

⁡

u

∂

⁡

t

2

⁢

∂

⁡

x

The notation

f

′

1

and

f

′

2

denotes the partial derivatives of the

function

f

with respect to its first and second

arguments

. Consider the following system of equations:

∂

⁡

u

∂

⁡

t

=

u

t

∂

⁡

u

t

∂

⁡

t

=

f

⁢

(

u

t

⁢

x

,

u

x

⁢

x

)

∂

⁡

u

x

∂

⁡

t

=

∂

⁡

u

t

∂

⁡

x

∂

⁡

u

t

⁢

t

∂

⁡

t

=

f

′

1

⁢

(

u

t

⁢

x

,

u

x

⁢

x

)

⁢

∂

⁡

u

t

⁢

x

∂

⁡

x

+

f

′

2

⁢

(

u

t

⁢

x

,

u

x

⁢

x

)

⁢

∂

⁡

u

t

∂

⁡

x

∂

⁡

u

t

⁢

x

∂

⁡

t

=

∂

⁡

u

t

⁢

t

∂

⁡

x

∂

⁡

u

x

⁢

x

∂

⁡

t

=

∂

⁡

u

t

⁢

x

∂

⁡

x

These equations will be satisfied if

u

satisfies

the orignal equation and the new variables are set equal to appropriate partial derivatives of

u

.

Conversely

, by eliminating variables, one can show that, for any solution of the above system of equations which satisfies the

boundary conditions

u

x

=

∂

⁡

u

∂

⁡

x

u

x

⁢

x

=

∂

2

⁡

u

∂

⁡

x

2

u

t

⁢

x

=

∂

⁡

u

t

∂

⁡

x

u

t

⁢

t

=

f

⁢

(

u

x

⁢

t

,

u

x

⁢

x

)

on the surface

t

=

0

, the variable

u

will satisfy the original equation.

The same approach can be used to rewrite systems of any number of equations in any number of variables involving derivatives of arbitrary order as a system involving only first derivatives such that the left hand side is the time derivative of a variable and the right hand side only involves spatial derivatives. However, trying to write out the solution in general can lead to awkward and complicated notation. There is no particularly good reason to do so because, upon seeing the example presented here, it becomes obvious that it is possible.

Title

example of eliminating higher-order derivatives

Canonical name

ExampleOfEliminatingHigherorderDerivatives

Date of creation

2013-03-22 14:37:10

Last modified on

2013-03-22 14:37:10

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

11

Author

rspuzio (6075)

Entry type

Example

Classification

msc 35A10

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