1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

370 Chapter 6 Laplace Transform

In general we may outline our procedure as follows:

Original

problem

L

Solution of

original problem

Transformed

problem

Solution of

transformed

problem

L−^1

In the step markedL−^1 , we must compute the function oftto which the solu-
tion of the transformed problem corresponds. This is the difficult part of the
process. The key property of the inverse transform is its linearity, as expressed
by

L−^1

(

c 1 F 1 (s)+c 2 F 2 (s)

)

=c 1 L−^1

(

F 1 (s)

)

+c 2 L−^1

(

F 2 (s)

)

.

This property allows us to break down a complicated transform into a sum of
simple ones.
A simple mass–spring–damper system leads to the initial value problem
u′′+au′+ω^2 u= 0 , u( 0 )=u 0 , u′( 0 )=u 1 ,

whose transform is

s^2 U−su 0 −u 1 +a(sU−u 0 )+ω^2 U= 0.

Determination ofUgives it as the ratio of two polynomials:

U(s)=su^0 +(u^1 +au^0 )

s^2 +as+ω^2

.

Although this expression is not in Table 2, it can be worked around to a func-
tion ofs+a/2, whose inverse transform is available. The shift theorem then
givesu(t).Thereis,however,abetterway.
The inversion of a rational function ofs(that is, the ratio of two polynomi-
als) can be accomplished by the technique of “partial fractions.” Suppose we
wishtocomputetheinversetransformof

U(s)=

cs+d

s^2 +as+b.

The denominator has two roots,r 1 andr 2 , which we assume for the moment
to be distinct. Thus

s^2 +as+b=(s−r 1 )(s−r 2 ),

U(s)= cs+d

(s−r 1 )(s−r 2 )

.