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FIGURE 9.3
Two-dimensional boundary when
ξ
is moved to the boundary.
Substitute the expression in Fig. 9.2 for
dS
into the above equation to obtain
u
dS
2
2
π
0
π/
2
1
1
lim
→
−
=
lim
→
−
u
2
d
θ
sin
ϕ
d
ϕ
2
0
0
S
0
2
u
2
π
=
lim
→
−
=−
2
π
u
(9.27)
2
0
For the two-dimensional case, utilize the geometry and notation of Fig. 9.3. Then the
integration on the augmented segment for the second integral in Eq. (9.23) appears as
0
q
π
q
ln
1
=−
0
()
θ
lim
→
dS
lim
→
ln
d
0
S
where the relationships ln
1
=−
ln
and
dS
=
d
θ
(Fig. 9.3) have been used. Note that
lim
→
0
ln
()
=
0
,
so this integration has no contribution to Eq. (9.23) when
ξ
moves to the
boundary.
Now consider the third integral on the right-hand side of Eq. (9.23). From the geometry
of Fig. 9.3, it can be seen that
i
a
i
=
1
cos
φ
+
2
sin
φ
=
where the expressions for
1
and
2
of Fig. 9.3 have been introduced. Hence
ln
1
u
∗
∂
q
∗
=
∂
n
=
∂
1
∂
∂
n
=−
i
a
i
2
=−
1
=−
2
=−
∂
n
0
u
π
0
u
π
0
1
uq
∗
dS
lim
→
=
lim
→
−
d
θ
=−
d
θ
=−
π
u
0
S
It is evident that the integrals contribute
−
π
u
to Eq. (9.23) as
→
0 for the two-dimensional
(
−
π
)
(
−
π
)
case. Substitute the contributions for the two
u
and three
2
u
dimensional problems
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