Integral Transformations - Heat Conduction: Mathematical Models and Analytical Solutions - page 470

also has a zero point s 2 of order l , the Heaviside expansion (B.27) will
have another form

If B ( s )

d s l − 1

e st

d l − 1 1

l A

( s )

lim
s

( s − s 2 ) .

( l − 1 ) ! B ( s )

→ s 2

In the expansion, of course, there are only n − ( m + l ) terms left corresponding to

the zero points of order 1.
Example 4.
Find f

s 1

f f

( t ) for ( s )=

1 . (2) Find f

( t ) for ( s )=

2 .

s 2

+ ( − )

s s 1

Solution.
1. s 2 + 1 has two zero points of order 1: ± i. Thus,

e st s = i + e st s = − i =

s
2 s

s
2 s

1
2 (

e − i t e i t f ( t )= + )= cos t .

2 has a zero point of order 1 and a zero point of order 2: s 1 =

2. s ( s − 1 ) 0and s 2 = 1.

Thus

e st
s

e st s = 0 +

1

e t

f ( t )= lim

s

= 1 + ( t − 1 ) .

] [ s ( s − 1 )

2

→ 1

Remark .If B ( s ) has only zero points of order 1, then

B ( s )=( s − s 1 )( s − s 2 ) ··· ( s − s n ) ,

B ( s k )=( s − s 1 )( s − s 2 ) ··· ( s − s k − 1 )( s − s k + 1 ) ··· ( s − s n ) | s = s k ,

( = , , ··· ) ( )

where s i i 1 2 n are the zero points of B s . This is quite useful for finding

some inverse image functions. For example,

1 1

4 e − t

1
4 e 3 t

f

If ( s )=

) ,

then f ( t )= − + .

( s + 1 )( s − 3

e − t
6 +

e 2 t
15 +

e − 3 t
10

1

f

If ( s )=

) ,

then f ( t )= − .

( s + 1 )( s − 2 )( s + 3

B.2.4 Convolution Theorem

The convolution of two functions f 1 ( t ) and f 2 ( t ) is

+ ∞

f 1 ( t ) ∗ f 2 ( t )= f 1 ( τ ) f 2 ( t − τ ) d τ .

− ∞

Since f 1 ( t ) ≡ 0and f 2 ( t ) ≡ 0for t < 0, we have

0 t

f 1 ( t ) ∗ f 2 ( t )= f 1 ( τ ) f 2 ( t − τ ) d τ + f 1 ( τ ) f 2 ( t − τ ) d τ

− ∞ 0

+ ∞ t

+ ( τ ) ( − τ )

τ =

( τ ) ( − τ )

τ . f 1 f 2 t d f 1 f 2 t d

t 0