Environmental Engineering Reference
In-Depth Information
Solution.
Based on the given boundary conditions, we have from Rows 1, 3 and 7
in Table 2.1,
m
2
l
sin
m
π
x
l
2
,
λ
m
=
,
X
m
(
x
)=
,
M
m
=
m
=
1
,
2
, ···
;
l
1
μ
n
a
2
sin
μ
n
y
a
a
2
sin2
μ
n
λ
=
,
X
n
(
y
)=
,
M
n
=
+
,
n
2
μ
n
y
ah
1
,
n
the
μ
n
are the positive zero points of
f
1
(
y
)=
tan
y
+
=
1
,
2
, ···
;
μ
k
b
2
sin
μ
k
z
b
+
ϕ
k
1
μ
k
b
2
sin2
(
μ
k
+
λ
k
=
,
X
k
(
z
)=
,
M
k
=
−
cos
2
ϕ
k
)
,
μ
k
2
z
bh
2
,
k
μ
k
are the positive zero points of
f
2
(
the
z
)=
tan
z
+
=
1
,
2
, ···
.
Thus the
W
ψ
(
x
,
y
,
z
,
t
)
is, by Eq. (6.84)
⎧
⎨
∞
∑
m
,
n
,
k
=
1
B
mnk
e
α
mnk
t
sin
u
=
W
ψ
(
x
,
y
,
z
,
t
)=
β
mnk
t
sin
μ
k
z
b
+
ϕ
k
sin
m
π
x
sin
μ
n
y
a
·
,
l
(6.89)
l
d
x
a
0
d
y
b
⎩
1
M
m
M
n
M
k
β
mnk
B
mnk
=
0
ψ
(
x
,
y
,
z
)
0
sin
μ
k
z
b
+
ϕ
k
d
z
sin
m
π
x
sin
μ
n
y
a
·
,
l
where
1
τ
0
+
m
2
B
2
μ
k
b
2
μ
2
1
2
l
n
a
α
mnk
=
−
+
+
,
4
m
2
A
2
μ
k
b
2
μ
n
a
2
1
2
l
2
mnk
β
mnk
=
+
+
−
4
α
.
Finally, the solution of PDS (6.88) 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
)
∂
t
t
+
W
ψ
(
x
,
y
,
z
,
t
)+
W
f
τ
(
x
,
y
,
z
,
t
−
τ
)
d
τ
.
0
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