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sin k
n
z
k
n
C
i cos k
n
z
0
!†
P
q
n
D
:
(6.126)
First of all we note that these functions satisfy both Eq. (
6.120
) and the boundary
condition (
6.122
)at
z
D
0. Moreover, substituting Eq. (
6.126
)forq
n
into the
boundary condition (
6.122
)at
z
D
l
1
we come to the identity taking into account
Eq. (
6.123
).
It can be shown that these eigenfunctions form a set of orthonormal functions in
the sense that
Z
l
1
q
n
.
z
/q
m
.
z
/d
z
D
ı
nm
;
(6.127)
0
where ı
nm
denotes the Kronecker symbol
1; n
D
m
I
0; n
¤
m:
ı
nm
D
(6.128)
To find the functions a
n
.x/ and b
n
.x/ we substitute Eq. (
6.119
)forE
x
and E
y
into Eqs. (
6.117
) and (
6.118
)
X
k
A
k
m
a
m
q
m
D
ik
y
X
m
b
0
m
ik
y
a
m
q
m
i
0
!J
x
;
(6.129)
m
X
ǚ
b
0
m
C
k
A
k
m
b
m
q
m
D
ik
y
X
m
a
0
m
q
m
i
0
!J
y
;
(6.130)
m
where the prime denotes derivative with respect to x.
Now we consider free oscillations of the electromagnetic field in the MHD
box. In other words, the sources of the driving/external currents and perturbations
are assumed to be “turn off” so that J
x
D
J
y
D
0. Multiplying both sides of
Eqs. (
6.129
) and (
6.130
)byq
n
.
z
/, integrating these equations over
z
from 0 to l
1
and using the orthogonality condition (
6.127
) we are thus left with the set
k
A
k
n
k
y
a
n
;
ik
y
b
0
n
D
(6.131)
ik
y
a
0
n
D
b
0
n
C
k
A
k
n
b
n
:
(6.132)
Eliminating a
n
from Eqs. (
6.131
) and (
6.132
) and rearranging, we obtain
k
y
k
A
0
b
0
n
k
A
k
n
k
A
k
n
k
y
C
k
A
k
n
k
y
b
n
D
0:
b
0
n
(6.133)
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