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where
q
m
=
k
y
Q
s
(
x
0
,z
)
Q
m
(
x
0
,z
)
.
(6.14)
d
d
x
ω
s
(
x
0
)
It is important that
T
0
=
0
. The vector equation (6.13) can be replaced
by the matrix differential equation
x
0
)
U
(
x
)=
T
(
x
)
U
(
x
)
.
(
x
−
(6.15)
Here
U
(
x
) denotes a matrix in which every vector column is the solution of
the system (6.13).
Field Pattern Close to FLR-Shell
First of all let us consider the particular case
T
(
x
)=
T
0
,
i.e.
T
1
=
T
2
=
...
=
0
.
and make change of variables
τ
=ln[
k
y
(
x
−
x
0
)]
that transforms (6.15) into equation with constant coecients
d
U
dt
=
T
0
U
.
A fundamental matrix of this equation can be written as
U
=exp(
T
0
τ
) = exp
{
T
0
[
k
y
(
x
−
x
0
)]
}
,
where, as usual,
exp(
T
0
τ
)=
1
+
T
0
τ
+
T
0
τ
2
2
+
···
.
In the general case, a solution of (6.15) can be presented in the form [63]:
U
(
x
)=
P
(
x
)exp
{
T
0
ln [
k
y
(
x
−
x
0
)]
}
,
(6.16)
where
x
0
)
2
P
2
+
···
is a regular matrix function. By virtue of the equality
T
0
=
0
,wehave
U
(
x
)=
P
0
+(
x
P
(
x
)=
P
0
+(
x
−
x
0
)
P
1
+(
x
−
x
0
)
2
P
2
+
−
x
0
)
P
1
+(
x
−
···
×{
1
+
T
0
ln [
k
y
(
x
−
x
0
)]
}
.
(6.17)
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