Cauchy initial value problem

Cauchy initial value problem

Let

D

be a subset of

ℝ

n

×

ℝ

,

(

x

0

,

t

0

)

a point of

D

, and

f

:

D

→

ℝ

be a function.

We say that a function

x

⁢

(

t

)

is a solution to the Cauchy (or initial value) problem

{

x

′

⁢

(

t

)

=

f

⁢

(

x

⁢

(

t

)

,

t

)

x

⁢

(

t

0

)

=

x

0

(1)

if

1.

x

is a

differentiable function

x

:

I

→

ℝ

n

defined on a

interval

I

⊂

ℝ

;

2.

one has

(

x

⁢

(

t

)

,

t

)

∈

D

for all

t

∈

I

and

t

0

∈

I

;

3.

one has

x

⁢

(

t

0

)

=

x

0

and

x

′

⁢

(

t

)

=

f

⁢

(

x

⁢

(

t

)

,

t

)

for all

t

∈

I

.

We say that a solution

x

:

I

→

ℝ

n

is a

maximal solution

if it cannot be extended to a bigger interval. More precisely given any other solution

y

:

J

→

ℝ

n

defined on an interval

J

⊃

I

and such that

y

⁢

(

t

)

=

x

⁢

(

t

)

for all

t

∈

I

, one has

I

=

J

(and hence

x

and

y

are the same function).

We say that a solution

x

:

I

→

ℝ

n

is a

global solution

if

D

⊂

=

ℝ

n

×

I

.

We say that a solution

x

:

I

→

ℝ

n

is

unique

if given any other solution

y

:

I

→

ℝ

n

one has

x

⁢

(

t

)

=

y

⁢

(

t

)

for all

t

∈

I

(i.e.

x

is the unique solution defined on the interval

I

).

0.1

Notation

Usually the

differential equation

in (1) is simply written as

x

′

=

f

⁢

(

x

,

t

)

.
Also, depending on the topics, the name chosen for the function and for the

variable

, can change. Other common choices are

y

′

=

f

⁢

(

y

,

t

)

or

y

′

=

f

⁢

(

y

,

x

)

.
It is also common to write

x

˙

=

f

⁢

(

x

,

t

)

when the independent variable represents a time value.

0.2

Examples

1.

The function

x

⁢

(

t

)

=

log

⁡

t

defined on

I

=

(

0

,

+

∞

)

is the unique maximal solution to the

Cauchy problem

:

{

x

′

⁢

(

t

)

=

1

/

t

x

⁢

(

1

)

=

0

.

In this case

f

⁢

(

x

,

t

)

=

1

/

t

,

D

=

{

(

x

,

t

)

:

t

≠

0

}

,

t

0

=

1

,

x

0

=

0

.

2.

The function

x

⁢

(

t

)

=

e

t

is a global (and hence maximal), unique solution to the Cauchy problem:

{

x

′

⁢

(

t

)

=

x

⁢

(

t

)

x

⁢

(

0

)

=

1

.

3.

Consider the Cauchy problem

{

x

′

⁢

(

t

)

=

3

2

⁢

x

3

x

⁢

(

0

)

=

0

.

The function

x

⁢

(

t

)

=

0

defined on

I

=

ℝ

is a global solution.
However the function

y

⁢

(

t

)

=

t

3

defined on

I

=

[

0

,

+

∞

)

is also a solution and so are the functions

z

⁢

(

t

)

=

{

(

t

-

c

)

3

if

t

≥

c

0

if

t

<

c

.

for every

c

≥

0

.
So there are no unique solutions. Moreover

y

is not a maximal solution.

Title

Cauchy initial value problem

Canonical

name

CauchyInitialValueProblem

Date of creation

2013-03-22 14:57:18

Last modified on

2013-03-22 14:57:18

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

14

Author

paolini (1187)

Entry type

Definition

Classification

msc 34A12

Synonym

Cauchy problem

Synonym

initial value problem

Related topic

InitialValueProblem

Related topic

DifferentialEquation

Related topic

CauchyKowalewskiTheorem

Defines

solution to the Cauchy problem

Defines

solution to the initial value problem

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