Environmental Engineering Reference
In-Depth Information
1
2
∂
∂
1
2
u
∂ˆ
∂
∂
x
μ
∂ˆ
1
3
˃ˆ
A
1
ˆ
=
x
(
u
ˆ)
+
x
−
x
+
∂
∂
1
2
∂
∂
1
2
v
∂ˆ
∂
∂
y
μ
∂ˆ
1
3
˃ˆ
A
2
ˆ
=
y
(
v
ˆ)
+
y
−
y
+
(2.38)
∂
∂
1
2
∂
1
2
w
∂ˆ
∂
∂
∂
z
μ
∂ˆ
1
3
˃ˆ
A
3
ˆ
=
z
(
w
ˆ)
+
z
−
z
+
∂
∂
and
v
s
.
We now show that each one-dimensional split operator
A
i
(
w
=
w
−
i
=
1
,
2
,
3
)
is positive
semidefinite (or positive definite if
˃>
0), cf. [
41
]. For simplicity, consider only the
case when domain
D
is a cube
[
0
,
X
]×[
0
,
Y
]×[
0
,
Z
]
. Then
X
X
X
∂ˆ
∂
2
1
2
ˆ
X
0
.
1
3
−
μˆ
∂ˆ
∂
2
dx
2
u
ˆ
A
1
ˆ
dx
=
˃ˆ
+
μ
dx
+
x
x
0
0
0
0 belongs to
S
−
,
Assume that
u
(
0
)>
0 and
u
(
X
)>
0. Then the boundary point
x
=
X
belongs to
S
+
. Applying condition (
2.8
)at
x
while point
x
=
=
0 and condition
(
2.7
)at
x
=
X
, we get
1
2
ˆ
X
0
=
−
μˆ
∂ˆ
∂
1
2
[
ˆ
2
u
2
2
(
X
)
u
(
X
)
+
ˆ
(
0
)
u
(
0
)
]≥
0
.
x
Since
˃>
0 and
μ>
0, we conclude that
Z
Y
X
(
A
1
ˆ,ˆ)
L
2
(
D
)
=
ˆ
A
1
ˆ
dxdydz
≥
0
.
0
0
0
In the same way one can show that
A
2
and
A
3
are also positive semidefinite
operators. It should be noted that this proof is also true for any region
D
which
represents a union of finite number of cubes.
On the other hand, the operator of the adjoint problem (
2.18
)-(
2.24
) and (
2.11
)
is the adjoint of
A
, and can be written as the sum
A
∗
=
A
1
+
A
2
+
A
3
where
1
2
∂
∂
1
2
u
∂
g
∂
∂
x
μ
∂
g
1
3
˃ˆ
A
1
g
=−
x
(
ug
)
−
x
−
x
+
∂
∂
1
2
∂
∂
1
2
v
∂
g
∂
∂
y
μ
∂
g
1
3
˃ˆ
A
2
g
=−
y
(
vg
)
−
y
−
y
+
(2.39)
∂
∂
2
∂
w
∂
∂
∂
z
μ
∂
1
1
2
g
g
∂
1
3
˃ˆ.
A
3
g
=−
z
(
)
−
z
−
z
+
wg
∂
∂
Suppose for simplicity that
μ
=
μ(
z
)
, and define the net functions on different
grids:
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