Environmental Engineering Reference
In-Depth Information
μ
∂ˆ
∂
ˆμ
∂
g
g
∇·
μ
∇
ˆ
dr
=
g
dS
−
dS
+
ˆ
∇·
μ
∇
gdr
,
n
∂
n
D
∂
D
∂
D
D
g
∇·
ˆ
s
dr
=
g
ˆ
s
·
n
dS
−
ˆ
∇·
g
s
dr
,
D
∂
D
D
where
g
s
=−
v
s
g
k
. Then
(
A
ˆ,
g
)
=
ˆ(
−
U
·∇
g
−∇·
μ
∇
g
+
˃
g
−∇·
g
s
)
dr
D
ˆμ
∂
g
μ
∂ˆ
∂
+
g
ˆ
U
·
n
dS
+
dS
−
g
dS
+
g
ˆ
s
·
n
dS
.
∂
n
n
∂
D
∂
D
∂
D
∂
D
D
into four integrals over
S
T
,
S
+
,
S
−
and
S
B
, and using conditions (
2.6
)-(
2.9
) and (
2.12
), we obtain that
Dividing the integrals over boundary
∂
(
A
ˆ,
g
)
=
ˆ(
−
U
·∇
g
−∇·
μ
∇
g
+
˃
g
−∇·
g
s
)
dr
D
provided that the function
g
satisfies the boundary conditions (
2.20
)-(
2.23
) (see
below). Thus, the Lagrange identity is fulfilled if
A
∗
g
=−
U
·∇
g
−∇·
μ
∇
g
+
˃
g
−∇·
g
s
.
On the other hand, multiplying Eq. (
2.4
)by
g
and integrating the result over the
space-time domain
D
×
(
0
,
T
)
, we get
g
N
drdt
T
T
T
g
∂ˆ
∂
drdt
+
gA
ˆ
drdt
=
Q
i
(
t
)ʴ(
r
−
r
i
)
.
t
i
=
1
0
D
0
D
0
D
Integration by parts of the first integral, together with conditions (
2.10
) and
g
(
r
,
T
)
=
0, leads to
T
T
g
∂ˆ
∂
ˆ
∂
g
0
drdt
=−
g
(
r
,
0
)ˆ
(
r
)
dr
−
drdt
t
∂
t
0
D
D
0
D
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