In a case of
0
0,
0
mm
=
=
=
,
CS
the equation (33) takes the following
form:
(2)
(2)
2
2
2
1
1
2ln2sin
()
(
)
()
,
2
2
2
i
i
CS
CS
d
K
g
d
g
−
+
−
−
==
where
()
2
2
2
12
2
12
2
2
2
1
.
sin
(cos
)
(cos
)
i
i
i
ch
g
dd
PP
dd
−
+
−
+
=
−
Let us obtain SIE that is defined on conical strips and make a numerical
experiment using the method of discrete singularities [26, 27]. Introducing the
definition
1
N
=−
+
,
()
()
0
22
,,
(1)
nm
mn
mn
yx
−
=−
,
(
)
1/
dl
=−
, systems (28), (29)
can be transformed to the following form:
()
01
1
2
,
1
1
,
:
;
im
in
mn
n
y
e
e
S
=−
=
()
()
1
22
,
,
1
1
1
(1
)
0,
:
.
()
in
mn
mn
n
n
y
e
CS
Nn
n
=−
−
=
+
(34)
After multiplication both parts of (34) on
1
i
e
and differentiation with respect
to
1
we obtain
()
()
1
22
,,
(1
)
0
in
mn
mn
n
n
ye
n
=−
−=,
1
CS
.
As coefficients are reduced after differentiation then we will add an additional
condition in case of
1
=
. Hence, the initial problem has reduced to the solution of
the system of linear equations with respect to
()
2
,
mn
y
:
()
01
1
2
,
,;
im
in
mn
n
y
e
e
S
=−
=
(35)
()
()
1
22
,
,
1
(1
)
0,
,
in
mn
mn
n
n
y
e
CS
n
=−
−
=
(36)
with the additional condition
()
()
22
,,
1
(1
)(1)
0
()
n
mn
mn
n
n
y
Nn
n
=−
−
−
=
+
.
(37)
Let us introduce the function
()
()
1
22
2
1
,
,
(
)
(1
)
in
mn
mn
n
n
ye
n
=−
=
−
,
(38)
when
1
,
−
. From (36) it follows that
2
1
1
(
)
0,
.
CS
=
(39)
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