Environmental Engineering Reference
In-Depth Information
4.4.2 Cylindrical Domain
We solve mixed problems in a cylindrical domain by using a cylindrical coordinate
system in which boundary conditions are separable.
Consider mixed problems in a cylinder
of radius
a
0
and height
H
:
x
2
y
2
Ω
+
<
a
0
,
consists of the cylindrical surface of radius
a
0
and
height
H
, the upper circle and the lower circle of radius
a
0
. If all combinations of
boundary conditions are considered on
0
<
z
<
H
. Its boundary
∂Ω
, there exist total 27 combinations. Their
corresponding complete and orthogonal sets of eigenfunctions are denoted by
∂Ω
u
ijk
(
,
θ
,
)=
(
)
Θ
(
θ
)
Z
k
(
)
,
=
,
,
,
=
,
=
,
, ···
.
r
z
R
i
r
z
i
1
2
3
j
1
k
1
2
9
j
Here
Θ
(
θ
)
are available in Table 4.1, and
R
i
(
r
)
and
Z
k
(
z
)
can be found from Ta-
j
ble 2.1.
Example 2.
Solve
⎨
a
2
u
t
+
τ
0
u
tt
=
Δ
u
(
r
,
θ
,
z
,
t
)+
F
(
r
,
θ
,
z
,
t
)
,
Ω
×
(
0
,
+
∞
)
,
L
(
u
,
u
r
)
|
r
=
a
0
=
0
,
u
|
z
=
0
=(
u
z
+
hu
)
|
z
=
H
=
0
,
(4.60)
⎩
u
(
r
,
θ
,
z
,
0
)=
Φ
(
r
,
θ
,
z
)
,
u
t
(
r
,
θ
,
z
,
0
)=
Ψ
(
r
,
θ
,
z
)
.
Solution.
The boundary conditions on the cylindrical surface
L
0 can
be of three kinds; PDS (4.60) thus includes three PDS, each corresponding to one
of three boundary conditions at
r
(
u
,
u
r
)
|
r
=
a
0
=
a
0
. Based on the given boundary conditions, we
use the eigenfunction sets in Row 3 in Table 2.1 and expand the solution for the case
Φ
=
=
F
=
0,
a
(
1
)
γ
mnl
t
cos
n
+
∞
∑
n
=
0
,
m
,
l
=
1
t
e
−
b
(
1
)
u
=
2
τ
0
mnl
cos
γ
mnl
t
+
mnl
sin
θ
a
(
2
)
γ
mnl
t
sin
n
J
n
(
sin
μ
l
z
b
(
2
)
+
mnl
cos
γ
mnl
t
+
mnl
sin
θ
k
mn
r
)
H
,
z
Hh
,
μ
l
are the positive zero-points of
f
where
(
z
)=
tan
z
+
2
μ
l
H
2
4
μ
(
n
)
k
mn
=
μ
(
n
)
1
2
m
a
0
m
a
0
.
τ
0
a
2
γ
mnl
=
λ
−
1
,
λ
=
+
,
τ
0
0 yields
a
(
1
)
a
(
2
)
0.
b
(
1
)
mnl
and
b
(
2
)
Applying the initial condition
u
|
t
=
0
=
mnl
=
mnl
=
mnl
can
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