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where
A
is the area of the cross-section and
S
is the boundary. The first term of (4) can be
written as
A
ω
∗
(
∇
2
A
ω
∗
ω
,ii
dA
(ω
∗
ω
,i
)
,i
−
ω
,i
ω
,i
]
dA
ω)
dA
=
=
[
A
∂
∂
dA
∂ω
∗
∂
dA
x
2
+
∂ω
∗
ω
∗
∂ω
∂
∂
ω
∗
∂ω
∂
x
2
∂ω
x
3
∂ω
=
+
−
x
2
x
2
∂
x
3
x
3
∂
∂
∂
x
3
A
A
(5)
Use the Gauss' integral theorem [Appendix II, Eq. (II.8)] on the first term of the right-hand
side of (5) to obtain
∂
∂
dA
ω
∗
∂ω
∂
∂
∂
ω
∗
∂ω
∂
A
(ω
∗
ω
S
ω
∗
ω
+
=
)
=
,i
dA
,i
a
i
dS
,i
x
2
x
2
x
3
x
3
A
S
ω
∗
∂ω
S
ω
∗
qdS
=
dS
=
i
=
2
,
3
(6)
∂
n
Since this is a two-dimensional problem, the volume
V
in Appendix II, Eq. (II.8) is changed
to area
A
and
S
now represents the boundary of the cross-section. The second term of the
right-hand side of (5) can be transformed as
∂ω
∗
∂
dA
x
2
+
∂ω
∗
x
2
∂ω
x
3
∂ω
A
ω
,i
ω
,i
dA
=
∂
∂
∂
x
3
A
2
ω
∗
)
A
(ω ω
,i
)
,i
dA
=−
A
ω(
∇
dA
+
2
ω
∗
)
q
∗
dA
=−
A
ω(
∇
+
S
ω
=
dA
i
2
,
3
(7)
where
q
∗
=
∂ω
∗
∂
. Gauss' theorem is used again in (7). Substitution of (6) and (7) into (5)
n
leads to
dS
n
−
ω
∂ω
∗
ω
∗
∂ω
∂
A
ω
∗
(
∇
2
2
ω
∗
)
ω)
dA
=
A
ω(
∇
dA
+
(8)
∂
n
S
Then (4) becomes
q
dS
ω
∗
−
ω
∂ω
∗
∂
2
ω
∗
)
A
ω(
∇
dA
+
=
0
(9)
n
S
ω
∗
,
a singular function, let
To find the fundamental solution
2
ω
∗
=−
∇
2
πδ(ξ
,x
)
(10)
where
and
x
are two points in the cross-section. The solution to (10) is (Eq. (9.22b) or
Haberman, 1987)
ξ
ln
1
r
ω
∗
=
(11)
which has the derivative, with respect to the outer normal of the cross-section (Eq. 9.23)
∂ω
∗
∂
1
r
2
n
=−
r
i
a
i
i
=
2
,
3
(12)
where
r
i
is the component of
r
in the
x
i
direction and
a
i
is the direction cosine of the
outer normal with respect to the
x
i
axis. Substitute (10), (11), and (12) into (9), along with
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