image equation

image equation

In solving an

initial value problem

leading to an

ordinary differential
equation

, the

Laplace transform

offers often a way to simplify the equation:
both sides are Laplace transformed.  The transformed equation, the so-called

image equation

, is in many cases simplier than the original

differential
equation

, since it does not contain the derivatives of the unknown

function

y

⁢

(

t

)

.  From the image equation one may solve the Laplace transform

Y

⁢

(

s

)

of

y

⁢

(

t

)

and then

inverse

transform

Y

⁢

(

s

)

getting

y

⁢

(

t

)

.

Let’s consider e.g. the ordinary

n

’th order linear
differential equation

a

0

⁢

d

n

⁢

y

d

⁢

t

+

a

1

⁢

d

n

-

1

⁢

y

d

⁢

x

n

-

1

+

…

+

a

n

-

1

⁢

d

⁢

y

d

⁢

t

+

a

n

⁢

y

⁢

(

t

)

=

f

⁢

(

t

)

(1)

subject to the

initial conditions

y

⁢

(

0

)

=

y

0

,

y

′

⁢

(

0

)

=

y

0

′

,

…

,

y

n

-

1

⁢

(

0

)

=

y

0

(

n

-

1

)

.

(2)

Due to the linearity of the Laplace transform the image equation
of (1) is

a

0

⁢

ℒ

⁢

{

d

n

⁢

y

d

⁢

x

n

}

+

a

1

⁢

ℒ

⁢

{

d

n

-

1

⁢

y

d

⁢

x

n

-

1

}

+

…

+

a

n

⁢

ℒ

⁢

{

y

⁢

(

t

)

}

=

ℒ

⁢

{

f

⁢

(

t

)

}

.

(3)

Denote

ℒ

{

y

(

t

)

}

=

:

Y

(

s

)

and

ℒ

{

f

(

t

)

}

=

:

F

(

s

)

.  We put into (3) the expressions
of the Laplace transforms of the derivatives on the left hand side (see “

Laplace transforms of derivatives

(

http://planetmath.org/LaplaceTransformsOfDerivatives

)”)
getting

a

0

⁢

[

s

n

⁢

Y

⁢

(

s

)

-

(

s

n

-

1

⁢

y

0

+

s

n

-

2

⁢

y

0

′

+

…

+

y

0

(

n

-

1

)

)

]

+

a

1

⁢

[

s

n

-

1

⁢

Y

⁢

(

s

)

-

(

s

n

-

2

⁢

y

0

+

s

n

-

3

⁢

y

0

′

+

…

+

y

0

(

n

-

2

)

)

]

+

…

…

…

…

+

a

n

-

1

⁢

[

s

⁢

Y

⁢

(

s

)

-

(

y

0

)

]

=

F

⁢

(

s

)

.

This equation is simplified to

(

a

0

⁢

s

n

+

a

1

⁢

s

n

-

1

+

…

+

a

n

-

1

⁢

s

+

a

n

)

⁢

Y

⁢

(

s

)

=

a

0

⁢

[

y

0

⁢

s

n

-

1

+

y

0

′

⁢

s

n

-

2

+

…

+

y

0

(

n

-

1

)

]

+

+

a

1

⁢

[

y

0

⁢

s

n

-

2

+

y

0

′

⁢

s

n

-

3

+

…

+

y

0

(

n

-

2

)

]

+

+

…

…

…

…

+

+

a

n

-

2

⁢

[

y

0

⁢

s

+

y

0

′

]

+

a

n

-

1

⁢

[

y

0

]

+

F

⁢

(

s

)

.

For brevity, denote in the last equation the polynomial
multiplier of

Y

⁢

(

s

)

by

φ

⁢

(

s

)

and the sum preceding

F

⁢

(

s

)

by

ψ

⁢

(

s

)

.  Then the equation can be written as

φ

⁢

(

s

)

⁢

Y

⁢

(

s

)

=

ψ

⁢

(

s

)

+

F

⁢

(

s

)

,

i.e.

Y

⁢

(

s

)

=

ψ

⁢

(

s

)

φ

⁢

(

s

)

+

F

⁢

(

s

)

φ

⁢

(

s

)

.

(4)

The function

Y

⁢

(

s

)

defined by (4) is the Laplace transform of
the solution

y

⁢

(

t

)

of the differential equation (1) which
satisfies the initial conditions (2).  If we now find a function

y

*

⁢

(

t

)

the Laplace transform of which is the function

Y

⁢

(

s

)

defined by (4), then

y

*

⁢

(

t

)

will do for

y

⁢

(

t

)

due to the
uniqueness property of Laplace transform expressed in the entry
“Mellin’s inverse formula (

http://planetmath.org/MellinsInverseFormula

)”.

If we seek the solution of (1) satisfying the

zero initial
conditions

x

0

=

x

0

′

=

x

0

′′

=

…

=

x

0

(

n

-

1

)

=

0

,

then

ψ

⁢

(

s

)

≡

0

and

Y

⁢

(

s

)

=

F

⁢

(

s

)

φ

⁢

(

s

)

,

i.e.

Y

⁢

(

s

)

=

F

⁢

(

s

)

a

0

⁢

s

n

+

a

1

⁢

s

n

-

1

+

…

+

a

n

.

Example.

The 4’th order differential equation

y

′′′′

⁢

(

t

)

+

y

⁢

(

t

)

=

0

(5)

should be solved with the initial conditions

y

⁢

(

0

)

=

y

′′′

⁢

(

0

)

=

1

,

y

′

⁢

(

0

)

=

y

′′

⁢

(

0

)

=

0

.

The image equation of (5) is

s

4

⁢

Y

⁢

(

s

)

-

s

3

⁢

y

⁢

(

0

)

-

s

2

⁢

y

′

⁢

(

0

)

-

s

⁢

y

′′

⁢

(

0

)

-

y

′′′

⁢

(

0

)

+

Y

⁢

(

s

)

=

0

,

i.e.

(

s

4

+

1

)

⁢

Y

⁢

(

s

)

=

s

3

+

1

.

Thus one needs to determine the inverse Laplace transform of

Y

⁢

(

s

)

=

1

4

⋅

4

⁢

s

3

s

4

+

1

+

1

s

4

+

1

.

(6)

The zeroes of the numerator

s

4

+

1

are the eighth roots of unity

e

i

⁢

π

4

,

e

i

⁢

3

⁢

π

4

,

e

i

⁢

5

⁢

π

4

,

e

i

⁢

7

⁢

π

4

, in other words the

complex numbers

±

1

±

i

2

.  By the special case (3) of
the

Heaviside formula

, the first addend of (6)
corresponds the original function

1

4

⁢

∑

±

e

±

1

±

i

2

⁢

t

=

e

t

2

+

e

-

t

2

2

⋅

e

i

⁢

t

2

+

e

-

i

⁢

t

2

2

=

cosh

⁡

t

2

⁢

cos

⁡

t

2

.

Utilizing also the general
Heaviside formula (

http://planetmath.org/HeavisideFormula

) (1), one can
get from (6) the result

y

⁢

(

t

)

:=

cosh

⁡

t

2

⁢

cos

⁡

t

2

+

1

2

⁢

(

cosh

⁡

t

2

⁢

sin

⁡

t

2

-

sinh

⁡

t

2

⁢

cos

⁡

t

2

)

.

References

1

N. Piskunov:

Diferentsiaal- ja
integraalarvutus kõrgematele tehnilistele
õppeasutustele. Teine köide.

Viies trükk.
Kirjastus Valgus, Tallinn (1966).

Title

image equation

Canonical name

ImageEquation

Date of creation

2014-03-20 20:16:56

Last modified on

2014-03-20 20:16:56

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

13

Author

pahio (2872)

Entry type

Topic

Classification

msc 34A05

Classification

msc 44A10

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