1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

6.2 Partial Fractions and Convolutions 373

The first term in this expression is recognized as the transform ofu 0 e−at.If
F(s)is a rational function, partial fractions may be used to invert the second
term. However, we can identify that term by solving the problem another way.
For example,

eat(u′+au)=eatf(t),

(

ueat

)′

=eatf(t),

ueat=

∫t

0

eat′f

(

t′

)

dt′+c,

u(t)=

∫t

0

e−a(t−t′)f

(

t′

)

dt′+ce−at.

The initial condition requires thatc=u 0. On comparing the two results, we
see that

L

[∫t

0

e−a(t−t′)f

(

t′

)

dt′

]

=^1

s+a

F(s).

Thus the transform of the combination ofe−atandf(t)on the left is the prod-
uct of the transforms ofe−atandf(t). This simple result can be generalized in
the following way.

Theorem 2.If g(t)and f(t)have Laplace transforms G(s)and F(s),respectively,
then

L

[∫t

0

g

(

t−t′

)

f

(

t′

)

dt′

]

=G(s)F(s). (3)

(This is known as the convolution theorem.)

The integral on the left is called theconvolutionofgandf, written

g(t)∗f(t)=

∫t

0

g

(

t−t′

)

f

(

t′

)

dt′.

It can be shown that the convolution follows these rules:

g∗f=f∗g, (4a)
f∗(g∗h)=(f∗g)∗h, (4b)
f∗(g+h)=f∗g+f∗h. (4c)
The convolution theorem provides an important device for inverting
Laplace transforms, which we shall apply to find the general solution of the
nonhomogeneous problem

u′′−au=f(t), u( 0 )=u 0 , u′( 0 )=u 1.