Environmental Engineering Reference
In-Depth Information
or
L
−
1
α
)
=
α
L
−
1
f
1
(
)
+
β
L
−
1
f
2
(
)
.
f
1
(
f
2
(
s
)+
β
s
s
s
Differential Property
Let
L
f
[
f
(
t
)] =
(
s
)
. Applying integration by parts leads to
L
f
(
)
=
s f
t
(
s
)
−
f
(
0
)
,
Re
(
s
)
>
k
.
In general,
L
f
(
n
)
(
s
n
f
s
n
−
1
f
s
n
−
2
f
(
f
(
n
−
1
)
(
t
)
=
(
s
)
−
(
0
)
−
0
)
−···−
0
)
,
Re
(
s
)
>
k
.
In particular, if
f
(
k
)
(
0
)=
0,
k
=
0
,
1
,
2
, ··· ,
n
−
1wehave
L
f
(
n
)
(
L
f
(
)
=
L
f
(
)
=
s f
s
2
f
s
n
f
t
(
s
)
,
t
(
s
)
,
··· ,
t
)
=
(
s
)
.
Integral Property
Suppose that
L
f
[
f
(
t
)] =
(
s
. Thus
L
t
0
)
d
t
1
s
f
f
(
t
)
=
(
s
)
.
)=
0
f
Proof.
Let
h
(
t
(
t
)
d
t
. Thus
h
(
t
)=
f
(
t
)
,
h
(
0
)=
0
.
h
(
By applying the differential property we obtain
L
[
t
)] =
sL
[
h
(
t
)]
so that
L
t
0
d
t
1
s
f
f
(
t
)
=
(
s
)
.
L
t
0
d
t
d
t
t
0
t
1
s
n
f
In general,
d
t
···
f
(
t
)
=
(
s
)
.
0
ntims
f
f
(
t
)
Similarly, we can prove that, if
L
[
f
(
t
)] =
(
s
)
and lim
t
exist,
t
→
0
L
f
tL
−
1
∞
s
d
s
∞
(
t
)
f
f
=
(
)
(
)=
(
)
.
s
d
s
or
f
t
s
t
s
Shifting Property
Let
L
f
f
e
α
t
f
[
f
(
t
)] =
(
s
)
.Then
L
[
(
t
)] =
(
s
−
α
)
,
Re
(
s
−
α
)
>
k
.
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