Environmental Engineering Reference
In-Depth Information
Finally, the solution of PDS (3.25) is
t
u
=
W
ϕ
(
x
,
t
)+
W
f
τ
(
x
,
t
−
τ
)
d
τ
0
+
∞
t
+
∞
e
−
(
x
−
ξ
)
2
e
−
(
x
−
ξ
)
2
1
2
a
√
π
1
2
a
√
π
f
(
ξ
,
τ
)
√
t
=
ϕ
(
ξ
)
4
a
2
t
d
ξ
+
d
τ
4
a
2
(
t
−
τ
)
d
ξ
.
t
−
τ
−
∞
0
−
∞
Remark 1.
When performing the inverse Fourier transformation, the following
properties are applied.
F
+
∞
−
∞
Convolution theorem.
F
[
f
1
(
x
)
∗
f
2
(
x
)] =
f
1
(
ξ
)
f
2
(
x
−
ξ
)
d
ξ
f
1
(
ω
)
f
2
(
ω
)
.
=
f
x
a
F
−
1
f
ω
)
=
1
|
Similarity property.
(
a
,
a
=
0
.
a
|
p
e
−
ω
2
F
e
−
px
2
Transformation formula.
=
,
Re
(
p
)
>
0
.
4
p
In particular, the similarity property and the transformation formula lead to
⎡
⎤
a
π
F
−
1
a
e
−
a
π
a
e
−
√
4
a
3
t
ω
2
2
F
−
1
e
−
(
ω
a
)
2
t
⎣
⎦
4
a
(
4
a
3
t
)
ω
F
−
1
=
=
4
a
1
√
4
a
2
x
2
4
a
2
t
1
2
a
√
π
x
2
4
a
2
t
t
e
−
t
e
−
=
=
.
π
Remark 2.
Let
u
in PDS (3.25) be temperature with
Θ
as its unit. Thus the unit of
Eq. (3.28) is
√
T
L
·
]=
a
−
1
2
1
√
T
·
·
T
·
1
−
τ
)
−
[
u
(
t
[
d
τ
][
f
(
ξ
,
τ
)] [
d
ξ
]=
T
L
=
Θ
,
where
2
−
(
x
−
ξ
)
=
1
.
4
a
2
(
t
−
τ
)
Therefore the unit in Eq. (3.28) is right. The unit is also correct in Eq. (3.27).
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