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Δ
u
(
x, y
) = 0
, x
∈
R
n
,
0
< y < a ,
in order to satisfy the boundary conditions
u
(
x,
0) =
ϕ
(
x
)
, x
∈
R
n
;
u
y
(
x, a
) =
ψ
(
x
)
, x
∈
R
n
.
Solution of the problem for the general case. Derivation of the
recurrence relation.
Let us apply the Fourier transform on
x
to the Laplace
equation, denoting
U
(
t, y
) =
F
x
[
u
] (
t, y
) ; Φ (
t
) =
F
[
ϕ
] (
t
) ; Ψ (
t
) =
F
[
ψ
](
t
)
.
We will obtain a boundary value problem for an ordinary differential
equation with a parameter
t
∈
R
n
:
− |
t
|
2
U
(
t, y
) +
U
yy
(
t, y
) = 0;
U
(
t,
0) = Φ (
t
)
, U
y
(
t, a
) = Ψ (
t
)
.
The solution to this boundary value problem is a function
U
(
t, y
) = Φ (
t
)
ch(
|
t
|
(
a
−
y
))
ch(
a
|
t
|
)
+ Ψ (
t
)
sh (
|
t
|
y
)
|
t
|
ch (
a
|
t
|
)
.
Applying the inverse Fourier transform, we are finding the solution to the
original mixed boundary value problem as a convolution
u
(
x, y
) =
ϕ
(
x
)
∗
K
n
(
x, y
) +
ψ
(
x
)
∗
L
n
(
x, y
)
,
(1)
where the kernels are designated as
K
n
(
x, y
) =
F
−
1
t
[
k
n
] (
x, y
)
, k
n
(
|
t
|
, y
) =
ch (
|
t
|
(
a
−
y
))
ch(
a
|
t
|
)
, t
∈
R
n
;
L
n
(
x, y
) =
F
−
1
t
[
l
n
] (
x, y
)
, l
n
(
|
t
|
, y
) =
sh (
|
t
|
y
)
|
t
|
ch (
a
|
t
|
)
, t
∈
R
n
.
Since
∂
∂y
l
n
(
|
t
|
, y
) =
k
n
(
|
t
|
, a
−
y
)
,
then
∂
∂y
L
n
(
x, y
) =
K
n
(
x, a
−
y
)
and
∂
∂y
L
n
(
x, y
)
as
y
→
a
−
0
, behaves in the same way as
K
n
(
x, y
)
if
y
→
+0
.
ISSN 1812-3368. Herald of the BMSTU. Series “Natural Sciences”. 2015. No. 1
5