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It is easy to test the feasibility of the following properties in the area
D
:
1)
K
n
(
x, y
)
>
0
;
2)
Z
R
n
K
n
(
x, y
)
dx
= 1
;
3) when
∀
δ >
0
,
lim
y
→
+0
sup
|
x
|≥
δ
K
n
(
x, y
) = 0
.
These properties mean that
K
n
(
x, y
)
is an
approximate identity
, or
δ
-shaped system of functions of
x
(with parameter
y
). When
y
→
+0
,
K
n
(
x, y
)
converges weakly to a
δ
-function
δ
(
x
)
. Since
∂
∂y
L
n
(
x, y
) =
K
n
(
x, a
−
y
)
,
∂
∂y
L
n
(
x, y
)
is an approximate identity as
y
→
a
−
0
.
Therefore, if
ϕ
(
x
)
∈ S
0
(
R
n
)
and
ψ
(
x
)
∈ S
0
(
R
n
)
, then the formula (1)
gives a generalized solution to the problem:
lim
y
→
+0
u
(
x, y
) =
ϕ
(
x
)
in
S
0
;
lim
y
→
a
−
0
u
y
(
x, y
) =
ψ
(
x
)
in
S
0
.
If the functions
ϕ
(
x
)
and
ψ
(
x
)
are normal functions of polynomial
growth, than at each point of continuity
lim
y
→
+0
u
(
x, y
) =
ϕ
(
x
) ; lim
y
→
a
−
0
u
y
(
x, y
) =
ψ
(
x
)
.
Solution to the mixed boundary value problem for the bandwidth
on the plane.
In the case of two variables in view of parity functions in
t
(
k
1
(
|
t
|
, y
) =
k
1
(
t, y
)
,
l
1
(
|
t
|
, y
) =
l
1
(
t, y
))
, the inverse Fourier transform
can be found from the tables [9]:
F
−
1
t
[
k
1
] (
x, y
) =
1
a
sin
πy
2
a
ch
πx
2
a
ch
πx
a
−
cos
πy
a
=
K
1
(
x, y
) ;
F
−
1
t
[
l
1
] (
x, y
) =
1
2
π
ln
ch
πx
2
a
+ sin
πy
2
a
ch
πx
2
a
−
sin
πy
2
a
=
L
1
(
x, y
)
.
If the functions
ϕ
(
x
)
and
ψ
(
x
)
are the functions of polynomial growth,
then the solution to the mixed problem is written as the integral formula
u
(
x, y
) =
1
a
sin
πy
2
a
∞
Z
−∞
ϕ
(
t
)
ch
π
(
x
−
t
)
2
a
ch
π
(
x
−
t
)
a
−
cos
πy
a
dt
+
ISSN 1812-3368. Herald of the BMSTU. Series “Natural Sciences”. 2015. No. 1
7