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In-Depth Information
d
2
U
(
λ,
z
,
z
o
)
2
(
−
λ,
z
)
U
(
λ,
z
,
z
o
)
=−
(
z
−
z
o
)
z
,
z
o
∈
[0
,
D]
dz
2
(10
.
37)
(
λ,
z
)
=
λ
2
−
i
o
N
(
z
)
e
>
0
with conditions
dU
(
λ,
z
,
z
o
)
+
λ
U
(
λ,
z
,
z
o
)
=
0
z
=
0
dz
dU
(
λ,
z
,
z
o
)
−
D
(
λ
)
U
(
λ,
z
,
z
o
)
=
0
z
=
D
dz
D
(
λ
)
=
λ
2
−
i
o
Re
>
0
.
D
D
Let us turn to (10.36) and find the asymptotics of the functions
h
y
(
y
,
M
o
)
and
h
z
(
y
,
M
o
)at
|
y
−
y
o
| →∞
.Givenlarge
|
y
−
y
o
|
, harmonics of low spatial
frequencies
λ
make the major contribution to
h
y
(
y
,
M
o
),
h
z
(
y
,
M
o
). Expanding
U
(
λ,
z
=
0
,
z
o
)inpowersofsmall
λ
, we get
λ
=
0
+
...,
dU
(
λ,
z
=
0
,
z
o
)
U
(
λ,
z
=
0
,
z
o
)
=
U
(
λ
=
0
,
z
=
0
,
z
o
)
+
λ
d
λ
whence, upon the substitution into (10.36) and integration, we obtain
O
i
o
U
(
λ
=
0
,
z
=
0
,
z
o
)
1
h
y
(
y
,
M
o
)
=
+
(
y
−
y
o
)
2
(
y
−
y
o
)
4
λ
=
0
+
O
(10
.
38)
2
i
o
1
dU
(
λ,
z
=
0
,
z
o
)
1
h
z
(
y
,
M
o
)
=
.
(
y
−
y
o
)
3
d
λ
(
y
−
y
o
)
5
H
A
H
A
In order to write the relations between
0
y
and
0
z
in the form containing the
MT impedance, we introduce the functions
λ
=
0
.
dU
(
λ,
z
,
z
o
)
V
y
(
z
)
=
U
(
λ
=
0
,
z
,
z
o
)
,
V
z
(
z
)
=
(10
.
39)
d
λ
0. The problem for
the function
V
z
(
z
) is solved by differentiating (10.37) with respect to
The function
V
y
(
z
) is the solution of problem (10.37) at
λ
=
λ
and setting
λ
=
0. Then,
d
2
V
z
(
z
)
dz
2
+
i
o
(
z
)
V
z
(
z
)
=
0
z
∈
[0
,
D]
z
=+
0
=−
dV
z
(
z
)
dz
V
y
(0)
.
(10
40)
z
=
D
−
dV
z
(
z
)
dz
−
i
o
D
V
z
(D)
=
0
.
