Environmental Engineering Reference
In-Depth Information
is
1
W
ϕ
(
τ
0
+
∂
B
2
W
(
λ
m
+
λ
n
+
λ
k
)
ϕ
(
=
,
,
,
)+
,
,
,
)
u
x
y
z
t
x
y
z
t
(6.79)
∂
t
t
+
W
ψ
(
,
,
,
)+
W
f
τ
(
,
,
,
−
τ
)
τ
,
x
y
z
t
x
y
z
t
d
(6.80)
0
where
f
τ
=
f
(
x
,
y
,
z
,
τ
)
.
Proof
. By Theorem 2 in Section 6.1, we only need to prove that the solution of
⎧
⎨
u
t
τ
0
+
B
2
∂
∂
A
2
u
tt
=
Δ
u
+
t
Δ
u
,
Ω
×
(
0
,
+
∞
)
,
(6.81)
⎩
L
(
u
,
u
n
)
|
∂Ω
=
0
,
u
(
x
,
y
,
z
,
0
)=
ϕ
(
x
,
y
,
z
)
,
u
t
(
x
,
y
,
z
,
0
)=
0
is
1
τ
W
ϕ
(
0
+
∂
B
2
W
(
λ
m
+
λ
n
+
λ
k
)
ϕ
(
u
=
x
,
y
,
z
,
t
)+
x
,
y
,
z
,
t
)
.
(6.82)
∂
t
Similar to the two-dimensional case, we expand the solution of PDS (6.77) into
=
m
,
n
,
k
T
mnk
(
t
)
X
m
(
x
)
Y
n
(
y
)
Z
k
(
z
)
.
u
(6.83)
Substituting this into the equation yields the
T
mnk
(
t
)
-equation
1
τ
0
+(
λ
m
+
λ
n
+
λ
k
)
B
2
T
mnk
(
T
mnk
(
A
2
T
mnk
(
t
)+
t
)+(
λ
m
+
λ
n
+
λ
k
)
t
)=
0
.
Its characteristic roots read
r
1
,
2
=
α
mnk
±
β
mnk
i
,
where
1
τ
0
+(
λ
B
2
4
1
2
1
2
2
mnk
α
mnk
=
−
+
λ
+
λ
k
)
,
β
mnk
=
(
λ
+
λ
+
λ
k
)
A
2
−
4
α
.
m
n
m
n
Thus the solution (6.83) becomes
=
m
,
n
,
k
e
α
mnk
t
(
+
)
(
)
(
)
Z
k
(
)
,
u
A
mnk
cos
β
mnk
t
B
mnk
sin
β
mnk
t
X
m
x
Y
n
y
z
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