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equations in first order form appear as
∂
z
r
=
Az
+
P
∂
with
z
and
P
the same as in Eq. (13.73) and
.
.
.
−
0
1
0
0
··················
··················
···
······
.
.
.
)
∂
∂φ
r
2
(ν/
−
ν/
r
0
1
/
K
2
··················
··················
···
······
K
(1
−
ν
2
.
.
.
2
)
4
∂φ
2
∂φ
∂
2(1
−
ν
)
∂
−
A
=
(13.74)
r
4
4
r
4
2
)
K
(
3
−
2
ν
−
ν
2
∂φ
2
∂φ
∂
)
∂
−
−
1
/
r
−
(ν/
r
n
φ
+
t
2
.
.
.
r
3
2
2
2
r
2
∂φ
∂φ
+
ρ
∂
1
+
k
∂
··················
··················
···
······
.
−
.
.
2
)
K
(1
−
ν
n
r
+
2
)
K
(3
−
2
ν
−
ν
2
∂φ
r
2
∂
−
1
−
(
1
−
ν)/
r
r
3
2
.
−
.
.
2
∂φ
2
K
(1
−
ν
)
r
2
∂
2
Reduction to Ordinary Differential Equations
The
derivatives can be eliminated in Eqs. (13.73) and (13.74) by expanding the variables
in a Fourier series
φ
w(
φ)
w
c
m
(
)
w
s
m
(
)
r,
r
r
=
∞
θ(
φ)
θ
m
(
)
θ
m
(
)
r,
r
r
z
=
cos
m
φ
+
sin
m
φ
(13.75)
V
(
r,
φ)
V
m
(
r
)
V
m
(
r
)
m
=
0
m
r
(
r,
φ)
m
c
m
(
r
)
m
s
m
(
r
)
The loading
p
z
(
r,
φ)
should also be expanded in a Fourier series
∞
p
c
m
(
φ
p
s
m
(
p
z
(
r,
φ)
=
r
)
cos
m
φ
+
r
)
sin
m
(13.76)
m
=
0
Multiply both sides of Eq. (13.76) by cos
n
φ
and integrate from
φ
=
0to
φ
=
2
π
,
i.e.,
2
π
2
π
2
π
∞
p
c
m
(
p
s
m
(
p
z
(
r,
φ)
cos
n
φ
d
φ
=
r
)
cos
m
φ
cos
n
φ
d
φ
+
r
)
sin
m
φ
cos
n
φ
d
φ
0
0
0
m
=
0
Since
2
π
2
π
2
π
m
=
n
=
0
cos
m
φ
cos
n
φ
d
φ
=
π
m
=
n
and
sin
m
φ
cos
n
φ
d
φ
=
0
0
0
m
=
n
0
it follows that
2
π
1
2
p
0
(
r
)
=
p
z
(
r,
φ)
d
φ
(13.77a)
π
0
2
π
1
π
p
c
m
(
r
)
=
p
z
(
r,
φ)
cos
m
φ
d
φ
m
>
0
(13.77b)
0
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