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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