Environmental Engineering Reference
In-Depth Information
Hence the
u
(
M
,
t
)
in Eq. (6.161) satisfies the equation of PDS (6.160). Clearly, the
u
(
M
,
t
)
in Eq. (6.161) also satisfies the initial condition
u
(
M
,
0
)=
0. Also,
t
t
W
f
τ
τ
=
t
,
)=
∂
∂
∂
W
f
τ
∂
u
t
(
M
,
t
W
f
τ
(
M
,
t
−
τ
)
d
τ
=
d
τ
+
t
t
0
0
which shows that
u
t
(
M
,
0
)=
0 by the initial condition of PDS (6.162). Therefore
the
u
(
M
,
t
)
in Eq. (6.161) is indeed the solution of PDS (6.160).
Remark 1
.Let
u
=
W
ψ
(
M
,
t
)
be the solution of PDS (6.156). The solution of
⎧
⎨
u
t
τ
0
+
B
2
∂
∂
A
2
u
tt
=
Δ
u
+
t
Δ
u
+
f
(
M
,
t
)
,
Ω
×
(
0
,
+
∞
)
,
(6.163)
⎩
u
(
M
,
0
)=
ϕ
(
M
)
,
u
t
(
M
,
0
)=
ψ
(
M
)
.
is, thus by the principle of superposition,
1
W
ϕ
(
t
τ
0
+
∂
B
2
u
=
t
−
Δ
M
,
t
)+
W
ψ
(
M
,
t
)+
W
f
τ
(
M
,
t
−
τ
)
d
τ
,
∂
0
where
f
τ
=
f
(
M
,
τ
)
.
Remark 2
.The
W
ψ
(
M
,
t
)
can be readily obtained by integral transformations. The
one-dimensional
W
ψ
(
can be obtained by using either the Laplace transforma-
tion with respect to the temporal variable
t
or the Fourier transformation with respect
to the spatial variable
x
. The two- or three-dimensional
W
ψ
(
M
,
t
)
can be found by
using multiple Fourier transformations. For the three-dimensional case, for example,
the
W
ψ
(
M
,
t
)
M
,
t
)
is the solution of
⎧
⎨
u
t
τ
0
+
B
2
∂
∂
A
2
R
3
u
tt
=
Δ
u
+
t
Δ
u
,
×
(
0
,
+
∞
)
,
(6.164)
⎩
u
(
M
,
0
)=
0
,
u
t
(
M
,
0
)=
ψ
(
M
)
.
Its triple Fourier transformation yields
⎧
⎨
1
τ
0
+
2
u
t
(
ω
,
B
2
A
2
2
u
t
(
ω
,
u
tt
(
ω
,
t
)+
ω
t
)+
ω
t
)=
0
,
(6.165)
⎩
¯
u
(
ω
,
0
)=
0
,
u
t
(
ω
,
0
)=
ψ
(
ω
)
.
Let
α
(
ω
)
±
i
β
(
ω
)
be the characteristic roots of the equation of PDS (6.165). Then
e
α
(
ω
)
t
u
(
ω
,
t
)=
[
A
(
ω
)
cos
β
(
ω
)
t
+
B
(
ω
)
sin
β
(
ω
)
t
]
,
where the
A
(
ω
)
and the
B
(
ω
)
are functions of
ω
to be determined. Applying the
initial conditions in (6.165) leads to
ψ
(
ω
)
β
(
ω
)
.
¯
A
(
ω
)=
0
,
B
(
ω
)=
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