Environmental Engineering Reference
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so that
1
τ
0
+(
λ
m
+
λ
n
+
λ
k
)
B
2
D
mnk
=
−
α
mnk
β
mnk
C
mnk
=
2
1
M
mnk
β
mnk
·
ϕ
(
x
,
y
,
z
)
X
m
(
x
)
Y
n
(
y
)
Z
k
(
z
)
d
x
d
y
d
z
.
(6.86)
Ω
By Eq. (6.84), we have
⎧
⎨
=
m
,
n
,
k
D
mnk
e
α
mnk
t
α
mnk
sin
β
mnk
t
+
β
mnk
cos β
mnk
t
·
∂
W
ϕ
(
x
,
y
,
z
,
t
)
∂
t
X
m
(
x
)
Y
n
(
y
)
Z
k
(
z
)
,
⎩
1
M
mnk
β
mnk
D
mnk
=
ϕ
(
x
,
y
,
z
)
X
m
(
x
)
Y
n
(
y
)
Z
k
(
z
)
d
x
d
y
d
z
.
Ω
Thus
m
,
n
,
k
C
mnk
e
α
mnk
t
cos β
mnk
t
·
X
m
(
x
)
Y
n
(
y
)
Z
k
(
z
)
1
τ
0
+(
λ
m
+
λ
n
+
λ
k
)
B
2
=
∂
∂
1
M
mnk
β
mnk
)+
m
,
n
,
k
t
W
ϕ
(
,
,
,
x
y
z
t
2
⎤
⎦
e
α
mnk
t
sin
·
ϕ
(
x
,
y
,
z
)
X
m
(
x
)
Y
n
(
y
)
Z
k
(
z
)
d
x
d
y
d
z
β
mnk
t
·
X
m
(
x
)
Y
n
(
y
)
Z
k
(
z
)
.
(6.87)
Ω
Substituting Eqs. (6.86) and (6.84) into Eq. (6.85) and using the structure of
W
ψ
(
x
,
y
,
z
,
t
)
in Eq. (6.84) yields the solution of PDS (6.81)
1
W
ϕ
(
τ
0
+
∂
B
2
W
u
(
x
,
y
,
z
,
t
)=
x
,
y
,
z
,
t
)+
(
λ
m
+
λ
n
+
λ
k
)
ϕ
(
x
,
y
,
z
,
t
)
.
∂
t
Example 3
.Solve
⎧
⎨
u
t
τ
0
+
B
2
∂
∂
A
2
u
tt
=
Δ
u
+
t
Δ
u
+
f
(
x
,
y
,
z
,
t
)
,
0
<
x
<
l
,
0
<
y
<
a
,
0
<
z
<
b
,
0
<
t
,
u
(
0
,
y
,
z
,
t
)=
u
(
l
,
y
,
z
,
t
)=
0
,
(6.88)
⎩
u
(
x
,
0
,
z
,
t
)=
u
y
(
x
,
a
,
z
,
t
)+
h
1
u
(
x
,
a
,
z
,
t
)=
0
,
(
,
,
,
)
−
(
,
,
,
)=
(
,
,
,
)=
,
u
z
x
y
0
t
h
2
u
x
y
0
t
u
x
y
b
t
0
u
(
x
,
y
,
z
,
0
)=
ϕ
(
x
,
y
,
z
)
,
u
t
(
x
,
y
,
z
,
0
)=
ψ
(
x
,
y
,
z
)
.
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