Environmental Engineering Reference
In-Depth Information
ˆμ
∂ˆ
∂
2
ˆ
∇·
(μ
∇
ˆ)
dr
=
dS
−
μ
|∇
ˆ
|
dr
,
n
D
∂
D
D
1
2
ˆ
∇·
ˆ
s
dr
=
ˆˆ
s
·
n
dS
.
D
∂
D
S
+
,
S
−
Finally, dividing each integral over
∂
D
into the four integrals over
S
T
,
and
S
B
, and applying the conditions (
2.6
)-(
2.9
) and (
2.12
), we get
2
dr
2
2
k
(
ˆ,ˆ)
=
˃ˆ
+
μ
|∇
ˆ
|
+
S
T
ʶˆ
·
A
dr
n
dS
D
D
n
dS
1
2
2
dS
2
dS
2
k
2
k
+
U
n
ˆ
−
U
n
ˆ
+
v
s
ˆ
·
−
v
s
ˆ
·
n
dS
S
+
S
−
S
T
S
B
(2.15)
0in
S
−
,
k
Since
U
n
<
·
n
>
0at
S
T
and
k
·
n
<
0at
S
B
,Eq.(
2.15
) can be
rewritten as
2
dr
2
dr
2
k
μ
|
∇
ˆ
|
(
A
ˆ,ˆ)
=
˃ˆ
+
+
S
T
ʶˆ
·
n
dS
D
D
⊧
⊨
⊫
⊬
1
2
2
dS
2
|
U
n
|
ˆ
|
|
+
+
v
s
ˆ
k
·
n
dS
.
⊩
⊭
S
+
∪
S
−
S
T
∪
S
B
Thus, operator
A
is positive semidefinite:
0.
Taking the inner product of every term of Eq. (
2.4
) with
(
A
ˆ,ˆ)
≥
ˆ
we obtain
∂ˆ
∂
N
t
,ˆ
=
(
f
,ˆ)
−
(
A
ˆ,ˆ),
f
(
r
,
t
)
=
Q
i
(
t
)ʴ(
r
−
r
i
).
i
=
1
Using the condition
(
A
ˆ,ˆ)
≥
0 and the Schwarz inequality [
17
], the last equation
implies the inequality
ˆ,
∂ˆ
∂
≤
ˆ
f
,
ˆ
=
(ˆ, ˆ).
t
Further,
ˆ,
∂ˆ
∂
1
2
∂
∂
∂
2
=
t
ˆ
=
ˆ
t
ˆ
t
∂
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