aarjaneiro/planetmath huggingface dataset:
bound on matrix differential equation

bound on matrix differential equation

Suppose that

A

and

Z

are two square matrices dependent
on a

parameter

which satisfy the

differential equation

Z

′

⁢

(

t

)

=

A

⁢

(

t

)

⁢

Z

⁢

(

t

)

withh

initial condition

Z

⁢

(

0

)

=

I

.
Letting

∥

⋅

∥

denote the matrix

operator norm

, we
will show that, if

∥

A

⁢

(

t

)

∥

≤

C

for some constant

C

when

0

≤

t

≤

R

, then

∥

Z

⁢

(

t

)

-

I

∥

≤

C

⁢

(

e

C

⁢

t

-

1

)

when

0

≤

t

≤

R

.

We begin by applying the product

inequality

for the norm,
then employing the

triangle inequality

(both in the sum and
integral forms) after expressing

Z

as the

integral

of its

derivative

:

∥

Z

′

⁢

(

t

)

∥

≤

∥

A

⁢

(

t

)

∥

⁢

∥

Z

⁢

(

t

)

∥

≤

C

⁢

∥

Z

⁢

(

t

)

∥

=

C

⁢

∥

I

+

∫

0

t

𝑑

s

⁢

Z

′

⁢

(

s

)

∥

≤

C

⁢

∥

I

∥

+

C

⁢

∫

0

t

𝑑

s

⁢

∥

Z

′

⁢

(

s

)

∥

≤

C

+

C

⁢

∫

0

t

𝑑

s

⁢

∥

Z

′

⁢

(

s

)

∥

For convenience, let us define

f

⁢

(

t

)

=

∫

0

t

𝑑

s

⁢

∥

Z

′

⁢

(

s

)

∥

.
Then we have

f

′

⁢

(

t

)

≤

C

+

C

⁢

f

⁢

(

t

)

according to the
foregoing

derivation

. By the

product rule

,

d

d

⁢

t

⁢

(

e

-

C

⁢

t

⁢

f

⁢

(

t

)

)

=

e

-

C

⁢

t

⁢

(

f

′

⁢

(

t

)

-

C

⁢

f

⁢

(

t

)

)

.

Since

f

′

⁢

(

t

)

-

C

⁢

f

⁢

(

t

)

≤

C

, we have

d

d

⁢

t

⁢

(

e

-

C

⁢

t

⁢

f

⁢

(

t

)

)

≤

C

⁢

e

-

C

⁢

t

.

Taking the integral from

0

to

t

of both sides and
noting that

f

⁢

(

0

)

=

0

, we have

e

-

C

⁢

t

⁢

f

⁢

(

t

)

≤

C

⁢

(

1

-

e

-

C

⁢

t

)

.

Multiplying both sides by

e

C

⁢

t

and recalling the

definition

of

f

, we conclude

∫

0

t

𝑑

s

⁢

∥

Z

′

⁢

(

s

)

∥

≤

C

⁢

(

e

C

⁢

t

-

1

)

.

Finally, by the triangle inequality,

∥

Z

⁢

(

t

)

-

I

∥

=

∥

∫

0

t

𝑑

s

⁢

Z

⁢

(

s

)

∥

≤

∫

0

t

𝑑

s

⁢

∥

Z

⁢

(

s

)

∥

.

Combining this with the inequality derived in the
last paragraph produces the answer:

∥

Z

⁢

(

t

)

-

I

∥

≤

C

⁢

(

e

C

⁢

t

-

1

)

.

Title

bound on matrix differential equation

Canonical

name

BoundOnMatrixDifferentialEquation

Date of creation

2013-03-22 18:59:00

Last modified on

2013-03-22 18:59:00

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

4

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 34A30

Generated on Fri Feb 9 20:40:46 2018 by

LaTeXML
Chebyshev equation

Chebyshev equation

Chebyshev’s equation is the

second order

linear differential equation

(

1

-

x

2

)

⁢

d

2

⁢

y

d

⁢

x

2

-

x

⁢

d

⁢

y

d

⁢

x

+

p

2

⁢

y

=

0

where

p

is a real

constant

.

There are two independent solutions which are given as series by:

y

1

⁢

(

x

)

=

1

-

p

2

2

!

⁢

x

2

+

(

p

-

2

)

⁢

p

2

⁢

(

p

+

2

)

4

!

⁢

x

4

-

(

p

-

4

)

⁢

(

p

-

2

)

⁢

p

2

⁢

(

p

+

2

)

⁢

(

p

+

4

)

6

!

⁢

x

6

+

⋯

and

y

2

⁢

(

x

)

=

x

-

(

p

-

1

)

⁢

(

p

+

1

)

3

!

⁢

x

3

+

(

p

-

3

)

⁢

(

p

-

1

)

⁢

(

p

+

1

)

⁢

(

p

+

3

)

5

!

⁢

x

5

-

⋯

In each case, the coefficients are given by the

recursion

a

n

+

2

=

(

n

-

p

)

⁢

(

n

+

p

)

(

n

+

1

)

⁢

(

n

+

2

)

⁢

a

n

with

y

1

arising from the choice

a

0

=

1

,

a

1

=

0

,
and

y

2

arising from the choice

a

0

=

0

,

a

1

=

1

.

The series

converge

for

|

x

|

<

1

; this is easy to see from the

ratio test

and the recursion

formula

above.

When

p

is a non-negative integer, one of these series will terminate,
giving a polynomial solution.
If

p

≥

0

is even, then the series for

y

1

terminates at

x

p

.
If

p

is odd, then the series for

y

2

terminates at

x

p

.

These polynomials are, up to

multiplication

by a constant, the

Chebyshev polynomials

. These are the only polynomial solutions of the

Chebyshev equation

.

(In fact, polynomial solutions are also obtained when

p

is a negative integer,
but these are not new solutions, since the Chebyshev equation is

invariant

under the substitution of

p

by

-

p

.)

Title

Chebyshev equation

Canonical name

ChebyshevEquation

Date of creation

2013-03-22 13:10:17

Last modified on

2013-03-22 13:10:17

Owner

mclase (549)

Last modified by

mclase (549)

Numerical id

6

Author

mclase (549)

Entry type

Definition

Classification

msc 34A30

Synonym

Chebyshev differential equation

Related topic

HermiteEquation

Generated on Fri Feb 9 20:40:48 2018 by

LaTeXML
equation y′′=f⁢(x)

equation

y

′′

=

f

⁢

(

x

)

A simple special case of the

second order linear differential equation with constant coefficients

is

d

2

⁢

y

d

⁢

x

2

=

f

⁢

(

x

)

(1)

where

f

is

continuous

.  We obtain immediately

d

⁢

y

d

⁢

x

=

C

1

+

∫

f

⁢

(

x

)

⁢

𝑑

x

,

y

=

C

1

⁢

x

+

C

2

+

∫

(

∫

f

⁢

(

x

)

⁢

𝑑

x

)

⁢

𝑑

x

.

(2)

A particular solution

y

⁢

(

x

)

of (1) satisfying the

initial conditions

y

⁢

(

x

0

)

=

y

0

,

y

′

⁢

(

x

0

)

=

y

0

′

is obtained more simply by integrating (1) twice between the limits (

http://planetmath.org/DefiniteIntegral

)

x

0

and

x

, thus getting

y

⁢

(

x

)

=

y

0

+

y

0

′

⋅

(

x

-

x

0

)

+

∫

x

0

x

(

∫

x

0

x

f

⁢

(

x

)

⁢

𝑑

x

)

⁢

𝑑

x

.

But here, the two first addends are the first terms of the

Taylor polynomial

of

y

⁢

(

x

)

,

expanded

by the powers of

x

-

x

0

, whence the

double integral

is the corresponding remainder term (

http://planetmath.org/RemainderVariousFormulas

)

∫

x

0

x

y

′′

⁢

(

x

)

⁢

(

x

-

t

)

⁢

𝑑

t

=

∫

x

0

x

f

⁢

(

t

)

⁢

(

x

-

t

)

⁢

𝑑

t

.

Hence the particular solution can be written with the simple

integral

as

y

⁢

(

x

)

=

y

0

+

y

0

′

⋅

(

x

-

x

0

)

+

∫

x

0

x

f

⁢

(

t

)

⁢

(

x

-

t

)

⁢

𝑑

t

.

(3)

The result may be generalised for the

n

th

order (

http://planetmath.org/ODE

)

differential equation

d

n

⁢

y

d

⁢

x

n

=

f

⁢

(

x

)

(4)

with corresponding

n

initial conditions:

y

⁢

(

x

)

=

y

0

+

y

0

′

⋅

(

x

-

x

0

)

+

y

0

′′

2

!

⁢

(

x

-

x

0

)

2

+

…

+

y

0

(

n

-

1

)

(

n

-

1

)

!

⁢

(

x

-

x

0

)

n

-

1

+

1

(

n

-

1

)

!

⁢

∫

x

0

x

f

⁢

(

t

)

⁢

(

x

-

t

)

n

-

1

⁢

𝑑

t

.

(5)

Title

equation

y

′′

=

f

⁢

(

x

)

Canonical name

EquationYFx

Date of creation

2013-03-22 18:35:33

Last modified on

2013-03-22 18:35:33

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Topic

Classification

msc 34A30

Classification

msc 34-01

Generated on Fri Feb 9 20:40:51 2018 by

LaTeXML
linear differential equation of first order

linear differential equation of first order

An ordinary

linear differential equation of first order

has the form

d

⁢

y

d

⁢

x

+

P

⁢

(

x

)

⁢

y

=

Q

⁢

(

x

)

,

(1)

where

y

means the unknown

function

,

P

and

Q

are two known

continuous functions

.

For finding the solution of (1), we may seek a function

y

which is

product

of two functions:

y

⁢

(

x

)

=

u

⁢

(

x

)

⁢

v

⁢

(

x

)

(2)

One of these two can be chosen freely; the other is determined according to (1).

We substitute (2) and the

derivative

d

⁢

y

d

⁢

x

=

u

⁢

d

⁢

v

d

⁢

x

+

v

⁢

d

⁢

u

d

⁢

x

in (1), getting

u

⁢

d

⁢

v

d

⁢

x

+

v

⁢

d

⁢

u

d

⁢

x

+

P

⁢

u

⁢

v

=

Q

,  or

u

⁢

(

d

⁢

v

d

⁢

x

+

P

⁢

v

)

+

v

⁢

d

⁢

u

d

⁢

x

=

Q

.

(3)

If we chose the function

v

such that

d

⁢

v

d

⁢

x

+

P

⁢

v

=

0

,

this condition may be written

d

⁢

v

v

=

-

P

⁢

d

⁢

x

.

Integrating here both sides gives

ln

⁡

v

=

-

∫

P

⁢

𝑑

x

or

v

=

e

-

∫

P

⁢

𝑑

x

,

where the

exponent

means an arbitrary

antiderivative

of

-

P

.  Naturally,

v

⁢

(

x

)

≠

0

.

Considering the chosen property of

v

in (3), this equation can be written

v

⁢

d

⁢

u

d

⁢

x

=

Q

,

i.e.

d

⁢

u

d

⁢

x

=

Q

⁢

(

x

)

v

⁢

(

x

)

,

whence

u

=

∫

Q

⁢

(

x

)

v

⁢

(

x

)

⁢

𝑑

x

+

C

=

C

+

∫

Q

⁢

e

∫

P

⁢

𝑑

x

⁢

𝑑

x

.

So we have obtained the solution

y

=

e

-

∫

P

⁢

(

x

)

⁢

𝑑

x

⁢

[

C

+

∫

Q

⁢

(

x

)

⁢

e

∫

P

⁢

(

x

)

⁢

𝑑

x

⁢

𝑑

x

]

(4)

of the given

differential equation

(1).

The result (4) presents the

general solution

of (1), since the arbitrary

C

may be always chosen so that any given

initial condition

y

=

y

0

when

x

=

x

0

is fulfilled.

Title

linear differential equation of first order

Canonical name

LinearDifferentialEquationOfFirstOrder

Date of creation

2013-03-22 16:32:09

Last modified on

2013-03-22 16:32:09

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

13

Author

pahio (2872)

Entry type

Derivation

Classification

msc 34A30

Synonym

linear ordinary differential equation of first order

Related topic

SeparationOfVariables

Generated on Fri Feb 9 20:40:56 2018 by

LaTeXML
separation of variables

separation of variables

Separation of variables

is a valuable tool for solving

differential equations

of the form

d

⁢

y

d

⁢

x

=

f

⁢

(

x

)

⁢

g

⁢

(

y

)

The above equation can be rearranged algebraically through

Leibniz notation

, treating dy and dx as

differentials

, to separate the

variables

and be conveniently

integrable

on both sides.

d

⁢

y

g

⁢

(

y

)

=

f

⁢

(

x

)

⁢

d

⁢

x

Instead of using differentials, we can also make use of the change of variables theorem for integration and the fact that if two integrable functions are

equivalent

, then their

primitives

differ by a

constant

C

. Here, we write

y

=

y

⁢

(

x

)

and

d

⁢

y

d

⁢

x

=

y

′

⁢

(

x

)

for clarity. The above equation then becomes:

y

′

⁢

(

x

)

g

⁢

(

y

⁢

(

x

)

)

=

f

⁢

(

x

)

Integrating both sides over

x

gives us the desired result:

∫

y

′

⁢

(

x

)

g

⁢

(

y

⁢

(

x

)

)

⁢

𝑑

x

=

∫

f

⁢

(

x

)

⁢

𝑑

x

+

C

By the change of variables theorem of integration, the left hand side is equivalent to an

integral

in the variable

y

:

∫

d

⁢

y

g

⁢

(

y

)

=

∫

f

⁢

(

x

)

⁢

𝑑

x

+

C

It follows then that

∫

d

⁢

y

g

⁢

(

y

)

=

F

⁢

(

x

)

+

C

where

F

⁢

(

x

)

is an antiderivative of

f

and

C

is the constant

difference

between the two primitives. This gives a general form of the solution. An explicit form may be derived by an initial value.

Example:

A population that is initially at

200

organisms increases at a rate of

15

%

each year. We then have a differential equation

d

⁢

P

d

⁢

t

=

P

+

0.15

⁢

P

=

1.15

⁢

P

The solution of this equation is relatively straightforward, we simply separate the variables algebraically and

integrate

.

∫

d

⁢

P

P

=

∫

1.15

⁢

𝑑

t

This is just

ln

⁡

P

=

1.15

⁢

t

+

C

or

P

=

C

⁢

e

1.15

⁢

t

When we substitute

P

⁢

(

0

)

=

200

, we see that

C

=

200

. This is where we get the general

relation

of exponential growth

P

⁢

(

t

)

=

P

0

⁢

e

k

⁢

t

Title

separation of variables

Canonical

name

SeparationOfVariables

Date of creation

2013-03-22 12:29:24

Last modified on

2013-03-22 12:29:24

Owner

slider142 (78)

Last modified by

slider142 (78)

Numerical id

9

Author

slider142 (78)

Entry type

Algorithm

Classification

msc 34A30

Classification

msc 34A09

Classification

msc 34A05

Related topic

LinearDifferentialEquationOfFirstOrder

Related topic

InverseLaplaceTransformOfDerivatives

Related topic

SingularSolution

Related topic

ODETypesReductibleToTheVariablesSeparableCase

Generated on Fri Feb 9 20:41:01 2018 by

LaTeXML
application of sine integral at infinity

application of sine integral at infinity

For finding the value of the improper integral

∫

0

∞

sin

⁡

a

⁢

x

x

⁢

(

1

+

x

2

)

d

x

:=

f

(

a

)

(

a

>

0

)

(1)

we first use the

partial fraction

representation (

http://planetmath.org/PartialFractionsOfExpressions

)

1

x

⁢

(

1

+

x

2

)

=

1

x

-

x

1

+

x

2

.

Thus we may write

f

⁢

(

a

)

=

∫

0

∞

sin

⁡

a

⁢

x

x

⁢

𝑑

x

-

∫

0

∞

x

⁢

sin

⁡

a

⁢

x

1

+

x

2

⁢

𝑑

x

.

But by the entry

sine integral at infinity

, the first integral equals

π

2

.  When we check

f

′

⁢

(

a

)

=

∫

0

∞

cos

⁡

a

⁢

x

1

+

x

2

⁢

𝑑

x

,

f

′′

⁢

(

a

)

=

-

∫

0

∞

x

⁢

sin

⁡

a

⁢

x

1

+

x

2

⁢

𝑑

x

,

we see that there is the

linear differential equation

f

⁢

(

a

)

=

π

2

+

f

′′

⁢

(

a

)

(2)

i.e.

f

′′

-

f

=

-

π

2

,

satisfied by the sought

function

a

↦

f

⁢

(

a

)

.  We have the

initial conditions

f

⁢

(

0

)

=

∫

0

∞

0

⁢

𝑑

x

=

0

,

f

′

⁢

(

0

)

=

∫

0

∞

d

⁢

x

1

+

x

2

=

/

0

∞

⁡

arctan

⁡

x

=

π

2

.

Therefore the

general solution

f

⁢

(

a

)

=

C

1

⁢

e

a

+

C

2

⁢

e

-

a

+

π

2

of (2) requires that

C

1

=

0

,

C

2

=

π

2

,  and consequently the sought

integral

f

⁢

(

a

)

has the value

∫

0

∞

sin

⁡

a

⁢

x

x

⁢

(

1

+

x

2

)

⁢

𝑑

x

=

π

2

⁢

(

1

-

e

-

a

)

(3)

Title

application of sine integral at infinity

Canonical name

ApplicationOfSineIntegralAtInfinity

Date of creation

2013-03-22 18:45:58

Last modified on

2013-03-22 18:45:58

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Application

Classification

msc 34A34

Classification

msc 34A12

Classification

msc 26A36

Classification

msc 26A24

Synonym

generalisation of sine integral at infinity

Generated on Fri Feb 9 20:41:07 2018 by

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differential equation of circles

differential equation of circles

All circles of the plane form a three-parametric family

(

x

-

a

)

2

+

(

y

-

b

)

2

=

r

2

.

The parametres

a

,

b

,

r

may be eliminated by using successive

differentiations

, when one gets

x

-

a

+

(

y

-

b

)

⁢

y

′

=

0

,

1

+

y

′

⁣

2

+

(

y

-

b

)

⁢

y

′′

=

0

,

3

⁢

y

′

⁢

y

′′

+

(

y

-

b

)

⁢

y

′′′

=

0

.

The two last equations allow to eliminate also

b

, yielding
the

differential equation

of all circles of the plane:

(

1

+

y

′

⁣

2

)

⁢

y

′′′

-

3

⁢

y

′

⁢

y

′′

⁣

2

=

0

It is of three, corresponding the number of parametres.

Title

differential equation of circles

Canonical name

DifferentialEquationOfCircles

Date of creation

2013-03-22 18:59:26

Last modified on

2013-03-22 18:59:26

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Example

Classification

msc 34A34

Classification

msc 51-00

Generated on Fri Feb 9 20:41:08 2018 by

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gradient system

gradient system

A

gradient system

in

ℝ

n

is an

autonomous

ordinary differential equation

x

˙

=

-

grad

⁡

V

⁢

(

x

)

(1)

defined by the

gradient

of

V

where

V

:

ℝ

n

→

ℝ

and

V

∈

C

∞

. The following results can be deduced from the definition of a gradient system.

Properties:

•

The

eigenvalues

of the

linearization

of (

1

) evaluated at

equilibrium point

are real.

•

If

x

0

is an isolated

minimum

of

V

then

x

0

is an

asymptotically stable

solution of (

1

)

•

If

x

⁢

(

t

)

is a solution of (

1

) that is not an equilibrium point then

V

⁢

(

x

⁢

(

t

)

)

is a strictly decreasing function and is

perpendicular

to the level curves of

V

.

•

There does not exists

periodic solutions

of (

1

).

References

HSD

Hirsch, W. Morris, Smale, Stephen, Devaney, L. Robert:

Differential Equations

,

Dynamical Systems

& An Introduction to Chaos. Elsevier Academic Press, New York, 2004.

Title

gradient system

Canonical name

GradientSystem

Date of creation

2013-03-22 15:14:25

Last modified on

2013-03-22 15:14:25

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

7

Author

Daume (40)

Entry type

Definition

Classification

msc 34A34

Generated on Fri Feb 9 20:41:10 2018 by

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Riccati equation

Riccati equation

The

nonlinear differential equation

d

⁢

y

d

⁢

x

=

f

⁢

(

x

)

+

g

⁢

(

x

)

⁢

y

+

h

⁢

(

x

)

⁢

y

2

(1)

is called

Riccati equation

.  If

h

⁢

(

x

)

≡

0

,  it is a question of a linear differential equation; if

f

⁢

(

x

)

≡

0

,  of a

Bernoulli equation

.  There is no general method for integrating explicitely the equation (1), but
via the

substitution

y

:=

-

w

′

⁢

(

x

)

h

⁢

(

x

)

⁢

w

⁢

(

x

)

one can convert it to a

homogeneous linear differential equation

with non-constant coefficients.

If one can find a particular solution

y

0

⁢

(

x

)

,  then one can easily verify that the substitution

y

:=

y

0

⁢

(

x

)

+

1

w

⁢

(

x

)

(2)

converts (1) to

d

⁢

w

d

⁢

x

+

[

g

⁢

(

x

)

+

2

⁢

h

⁢

(

x

)

⁢

y

0

⁢

(

x

)

]

⁢

w

+

h

⁢

(

x

)

=

0

,

(3)

which is a

linear differential equation of first order

with respect to the function

w

=

w

⁢

(

x

)

.

Example.

The Riccati equation

d

⁢

y

x

=

3

+

3

⁢

x

2

⁢

y

-

x

⁢

y

2

(4)

has the particular solution

y

:=

3

⁢

x

.  Solve the equation.

We substitute

y

:=

3

⁢

x

+

1

w

⁢

(

x

)

to (4), getting

d

⁢

w

d

⁢

x

-

3

⁢

x

2

⁢

w

-

x

=

0

.

For solving this first order equation (

http://planetmath.org/LinearDifferentialEquationOfFirstOrder

) we can put

w

=

u

⁢

v

,

w

′

=

u

⁢

v

′

+

u

′

⁢

v

,  writing the equation as

u

⋅

(

v

′

-

3

⁢

x

3

⁢

v

)

+

u

′

⁢

v

:=

x

,

(5)

where we choose the value of the

expression

in parentheses equal to 0:

d

⁢

v

d

⁢

x

-

3

⁢

x

2

⁢

v

=

0

After

separation of variables

and integrating, we obtain from here a solution

v

=

e

x

3

,  which is set to the equation (5):

d

⁢

u

d

⁢

x

⁢

e

x

3

=

x

Separating the

variables

yields

d

⁢

u

=

x

e

x

3

⁢

d

⁢

x

and integrating:

u

=

C

+

∫

x

⁢

e

-

x

3

⁢

𝑑

x

.

Thus we have

w

=

w

⁢

(

x

)

=

u

⁢

v

=

e

x

3

⁢

[

C

+

∫

x

⁢

e

-

x

3

⁢

𝑑

x

]

,

whence the

general solution

of the Riccati equation (4) is

y

=

3

⁢

x

+

e

-

x

3

C

+

∫

x

⁢

e

-

x

3

⁢

𝑑

x

.

It may be proved that if one knows three different solutions of Riccati equation (1), the each other solution may be expresses as a rational function of them.

Title

Riccati equation

Canonical

name

RiccatiEquation

Date of creation

2013-03-22 18:05:43

Last modified on

2013-03-22 18:05:43

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Result

Classification

msc 34A34

Classification

msc 34A05

Synonym

Riccati differential equation

Related topic

BernoulliEquation

Generated on Fri Feb 9 20:41:16 2018 by

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autonomous system

autonomous system

A

system of ordinary differential equation

is

autonomous

when it does not depend on time

(does not depend on the

independent variable

)

i.e.

x

˙

=

f

⁢

(

x

)

. In contrast

nonautonomous

is when the system of ordinary differential equation does depend on time

(does depend on the independent variable)

i.e.

x

˙

=

f

⁢

(

x

,

t

)

.

It can be noted that every nonautonomous system can be converted to an autonomous system by adding a

dimension

. i.e. If

𝐱

˙

=

𝐟

⁢

(

𝐱

,

t

)

𝐱

∈

ℝ

n

then it can be written as an autonomous system with

𝐱

∈

ℝ

n

+

1

and by doing a substitution with

x

n

+

1

=

t

and

x

˙

n

+

1

=

1

.

Title

autonomous system

Canonical

name

AutonomousSystem

Date of creation

2013-03-22 13:37:26

Last modified on

2013-03-22 13:37:26

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

6

Author

Daume (40)

Entry type

Definition

Classification

msc 34A99

Synonym

autonomous

Synonym

autonomous equation

Synonym

nonautonomous

Synonym

nonautonomous equation

Related topic

TimeInvariant

Related topic

SystemDefinitions

Defines

nonautonomous system

Generated on Fri Feb 9 20:41:28 2018 by

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Green’s function for differential operator

Green’s function for differential operator

Assume we are given

g

∈

𝒞

0

⁢

(

[

0

,

T

]

)

and we want to find

f

∈

𝒞

1

⁢

(

[

0

,

T

]

)

such that

{

f

′

⁢

(

t

)

=

g

⁢

(

t

)

f

⁢

(

0

)

=

0

(1)

Expression (

1

) is an example of

initial value problem

for an

ordinary differential equation

. Let us
show, that (

1

) can be put into the framework of the definition for Green’s function.

1.

Ω

x

=

Ω

y

=

[

0

,

T

]

.

2.

ℱ

⁢

(

Ω

x

)

=

{

f

∈

𝒞

1

⁢

(

[

0

,

T

]

)

|

f

⁢

(

0

)

=

0

}

𝒢

⁢

(

Ω

y

)

=

𝒞

0

⁢

(

[

0

,

T

]

)

.

3.

A

⁢

f

=

f

′

Thus (

1

) can be written as an

operator

equation

A

⁢

f

=

g

.

(2)

To find the Green’s function for (

2

) we proceed as follows:

f

⁢

(

t

)

=

δ

t

⁢

(

A

-

1

⁢

g

)

=

∫

0

t

g

⁢

(

t

′

)

⁢

𝑑

t

′

=

∫

0

T

G

⁢

(

t

,

t

′

)

⁢

g

⁢

(

t

′

)

⁢

𝑑

t

′

,

where

G

⁢

(

t

,

t

′

)

has the following form:

G

⁢

(

t

,

t

′

)

=

{

1

,

0

≤

t

≤

t

′

0

,

t

′

<

t

≤

T

(3)

Thus, function (

3

) is the Green’s function for the operator equation (

2

) and
then for the problem (

1

).

Its graph is presented in Figure

1

.

Figure 1:

The Green’s function for the problem (

1

).

Title

Green’s function for

differential operator

Canonical name

GreensFunctionForDifferentialOperator

Date of creation

2013-03-22 14:43:39

Last modified on

2013-03-22 14:43:39

Owner

mathforever (4370)

Last modified by

mathforever (4370)

Numerical id

7

Author

mathforever (4370)

Entry type

Example

Classification

msc 34A99

Classification

msc 34A30

Generated on Fri Feb 9 20:41:30 2018 by

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separating solution

separating solution

A

separating solution

of a smooth

planar dynamical system

is an oriented smooth curve such that

•

the curve is composed of trajectories of the

dynamical system

,

•

the curve does not go through any

equilibrium point

,

•

the curve is the boundary of some region in the plane.

The orientation of the curve is the same as the orientation of the solutions. In

addition

the curve does not need to be connected.

[

KGA

]

References

KGA

Khovanskiĭ, A.G.: Fewnomials.

American Mathematical Society

, Providence, 1991.

Title

separating solution

Canonical

name

SeparatingSolution

Date of creation

2013-03-22 14:06:17

Last modified on

2013-03-22 14:06:17

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

10

Author

Daume (40)

Entry type

Definition

Classification

msc 34A99

Classification

msc 34-00

Generated on Fri Feb 9 20:41:32 2018 by

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eigenfunction

eigenfunction

Consider the Sturm-Liouville system given by

d

d

⁢

x

⁢

[

p

⁢

(

x

)

⁢

d

⁢

y

d

⁢

x

]

+

q

⁢

(

x

)

⁢

y

+

λ

⁢

r

⁢

(

x

)

⁢

y

=

0

a

≤

x

≤

b

(1)

a

1

⁢

y

⁢

(

a

)

+

a

2

⁢

y

′

⁢

(

a

)

=

0

,

b

1

⁢

y

⁢

(

b

)

+

b

2

⁢

y

′

⁢

(

b

)

=

0

,

(2)

where

a

i

,

b

i

∈

ℝ

with

i

∈

{

1

,

2

}

and

p

⁢

(

x

)

,

q

⁢

(

x

)

,

r

⁢

(

x

)

are

differentiable functions

and

λ

∈

ℝ

. A non zero solution of the system defined
by (

1

) and (

2

) exists in general for a
specified

λ

. The

functions

corresponding to that
specified

λ

are called eigenfunctions.

More generally, if

D

is some linear

differential operator

, and

λ

∈

ℝ

and

f

is a function such that

D

⁢

f

=

λ

⁢

f

then we say

f

is an eigenfunction of

D

with
eigenvalue

λ

.

Title

eigenfunction

Canonical name

Eigenfunction

Date of creation

2013-03-22 12:48:00

Last modified on

2013-03-22 12:48:00

Owner

tensorking (373)

Last modified by

tensorking (373)

Numerical id

8

Author

tensorking (373)

Entry type

Definition

Classification

msc 34B24

Synonym

characteristics function

Defines

solution of system

Generated on Fri Feb 9 20:41:33 2018 by

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