here
N
=
,
()
()
()
1
1
,
,
1
,
1
mn
mn
mn
=−
−
.
Let us consider the function
(1)
1,
()
,
,
in
mn
n
ye
+
=−
=
−
, (12)
Unknown coefficients
(1)
,
mn
y
are Fourier coefficients of the function
()
1
(12) and can be determined by the formula
(1)
,1
1
()
2
in
mn
y
e
d
−
−
=
, (13)
Therefore, the solution of the given first boundary value problem is reduced to
the problem of finding the function
()
1
. Let us find such equation that the required
function
()
1
satisfies to it. For this purpose we will use the equation (10), substitute
integral representation for
,
mn
y
(13) in (10) and obtain
(1)
,1
1
()
2
in
mn
S
y
e
d
−
=
,
taking into account that
()
1
0,
CS
=
according to (11). As a result we come to
the singular integrated equation (SIE) for
1
, when
S
:
(
)
0
1
(1)
1
1
(
)
()
2
1
(
)
()
,
2
S
im
m
Fd
K
d
e
−
+
+
−
−
=
(14)
where:
()
0
1
()
()
in
n
n
F
e
Nn
n
−
−=
+
;
(15)
(1)
(
)
,
0
1
()
()
in
mn
n
n
K
e
Nn
n
−
−=
+
; (16)
(
)
(1)
(1)
,0
1
1
()
m
mn
n
n
Nn
n
=
=
−
+
.
Function
()
F
−
(15) defines a singular part of integral equation (14), and
()
K
−
(16) is the kernel of a regular part of this equation.
In the particular case when
0
=, SIE (14) has the following form:
0
1
1
1
1
ˆ
ln2sin
()
(
)
()
,
2
2
2
im
m
S
S
d
K
d
e
−
+
−
=−
(1)
ˆ
,
(
)
(
)
m
m
S
K
K
−
=
−
−
.
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