Environmental Engineering Reference
In-Depth Information
Therefore
u
(
x
,
y
,
0
)=
ϕ
(
x
,
y
)
. By taking the derivative of Eq. (5.72) with respect
to
t
, we obtain
t
2
D
At
2
2
∂
η
=
∂
+
∂
∂
K
(
ϕ
)
d
ξ
d
t
2
(
ψ
ch
)
·
2
π
At
t
(
ψ
ch
)
·
4
π
A
.
∂
∂
)
t
=
0
=
∂
∂
Thus, by noting that
ψ
ch is an even function of
t
so that
t
(
ψ
ch
0,
t
2
C
At
2
∂
K
(
ϕ
)
d
ξ
d
η
|
t
=
0
=
0. Taking the derivative of Eq. (5.71) with respect to
t
∂
yields
τ
0
1
1
∂
∂
t
t
e
−
e
−
u
t
=
−
2
K
(
ϕ
)
d
ξ
d
η
+
2
τ
0
K
(
ϕ
)
d
ξ
d
η
0
8
π
A
τ
4
π
A
τ
t
0
D
At
D
At
t
2
D
At
2
1
∂
∂
1
∂
t
2τ
0
t
2τ
0
τ
0
e
−
A
e
−
−
K
(
ϕ
)
d
ξ
d
η
+
K
(
ϕ
)
d
ξ
d
η
4
π
A
t
2
π
∂
D
At
τ
0
t
2
D
At
2
1
1
∂
t
t
e
−
A
e
−
=
−
2
K
(
ϕ
)
d
ξ
d
η
+
2
τ
0
K
(
ϕ
)
d
ξ
d
η
.
0
8
π
A
τ
2
π
∂
D
At
Therefore,
⎡
τ
0
1
⎣
−
t
e
−
u
t
|
t
=
0
=
K
(
ϕ
)
d
ξ
d
η
2
0
8
π
A
τ
D
At
⎤
t
=
0
t
2
D
At
2
1
∂
t
2τ
0
⎦
A
e
−
+
K
(
ϕ
)
d
ξ
d
η
=
0
.
2
π
∂
3.
The
u
in Eq. (5.70)
It is straightforward to show
u
(
x
,
y
,
0
)=
0. By taking deriva-
tives of Eq. (5.70) with respect to
t
, we obtain
⎧
⎨
⎫
⎬
⎡
⎣
⎤
⎦
t
τ
0
τ
0
1
∂
∂
t
−
τ
2
t
−
τ
2
e
−
e
−
u
t
=
K
(
f
)
d
ξ
d
η
d
τ
+
K
(
f
)
d
ξ
d
η
|
τ
=
t
.
⎩
⎭
2
π
A
t
0
D
A
(
t
−
τ
)
D
A
(
t
−
τ
)
Therefore,
u
t
(
x
,
y
,
0
)=
0.
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