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Substituting these relations into (5.60), we obtain the equation for
b
k
y
d
b
ωn
d
x
+
L
ωn
−
d
d
x
1
L
ωn
b
ωn
=
L
−
1
ωn
φ
ωn
(5.63)
L
ωn
with boundary conditions
x
=0;
l
x
d
b
ωn
d
x
=0
,
(5.64)
where
d
v
L
−
1
ωn
f
xωn
d
x
ik
y
B
0
L
−
1
ωn
f
yωn
.
Find
ξ
⊥n
(
x, t
)and
b
n
(
x, t
) by applying the inverse Laplace transform to
ξ
⊥ωn
(
x
)and
b
ωn
(
x
). For example, the longitudinal magnetic component is
given by
φ
ωn
=
B
0
−
+
∞
+
iσ
0
1
2
π
b
(
x, t
)=
b
ωn
(
x
)
·
exp (
−
iωt
)d
ω
(5.65)
−∞
+
iσ
0
and similar expressions for displacements
ξ
⊥
.
In (5.65)
σ
0
>
0, the integration
path
Γ
(
+
iσ
0
) passes above all the peculiarities of function
b
ωn
(
x
) in the complex plane
ω
.
Find the solution of (5.63) with the right-hand side equal to
δ
(
x
−∞
+
iσ
0
,
+
∞
x
), i.e.
find the Green function
G
ωn
(
x, x
) of the boundary problem (5.63), (5.64).
For an arbitrary
φ
ωn
(
x
), perturbations
b
ωn
(
x
) are determined by the Green
integral
−
l
x
b
ωn
(
x
)=
G
ωn
(
x, x
)
·
f
ωn
(
x
)d
x.
0
According to the general theory
G
ωn
(
x, x
) can be expressed in terms of
two linearly independent solutions
ϕ
(1)
ωn
(
x
)and
ϕ
(2)
ωn
(
x
) of the uniform equation
(5.63). The first solution satisfies boundary condition (5.64) at
x
=0andthe
other at
x
=
l
x
. The following formula is valid:
ϕ
(1)
ωn
(
x
)
ϕ
(2)
ωn
(
x
)
for
x<x,
for
x>x,
G
ωn
(
x, x
)=
J
−
1
(5.66)
ωn
ϕ
(1)
ωn
(
x
)
ϕ
(2)
ωn
(
x
)
where
J
ωn
=
W
ωn
(
x
)
L
ωn
(
x
)
,
and it is independent of
x.
ωn
(
x
)
d
ϕ
(2)
ωn
(
x
)
d
ϕ
(1)
ωn
(
x
)
d
x
ωn
(
x
)
d
x
W
ωn
(
x
)=
ϕ
(1)
ϕ
(2)
−
is Wronskian of functions
ϕ
(1)
ωn
(
x
)and
ϕ
(2)
ωn
(
x
)
.
Consider the evolution of the initial perturbation in the simple case of
k
y
= 0, when Alfven and FMS-waves do not interact and we have uncoupled
equations.
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