Environmental Engineering Reference
In-Depth Information
By a variable transformation
x
=
cos
θ
, the equation in (7.22) is transformed into a
Legendre equation
d
2
Θ
d
x
2
−
2
x
d
d
x
+
λΘ
=
x
2
(
1
−
)
0
.
Its eigenvalues and eigenfunctions are available in Section 2.6,
λ
=
n
(
n
+
1
)
,
n
=
0
,
1
,
2
, ··· ,
Θ
n
(
θ
)=
P
n
(
cos
θ
)
.
Substituting
λ
=
n
(
n
+
1
)(
n
=
0
,
1
,
2
, ···
)
into the Euler equation (7.23) yields
r
2
R
(
2
rR
(
r
)+
r
)
−
n
(
n
+
1
)
R
(
r
)=
0
.
Its general solution is
a
n
r
n
b
n
r
−
(
n
+
1
)
,
R
n
(
r
)=
+
n
=
0
,
1
,
2
, ··· .
Since
|
R
n
(
0
)
| <
∞
, we obtain
b
n
=
0. Thus
∞
n
=
0
a
n
r
n
P
n
(
cos θ
)
,
(
,
θ
)=
≤
θ
≤
π
.
u
r
0
(7.24)
Note that
{
P
n
(
cos
θ
)
}
is orthogonal in
[
0
,
π
]
with respect to the weight function
sin
θ
. Applying the boundary condition in PDS (7.21) yields
π
2
n
+
1
a
n
=
f
(
θ
)
P
n
(
cos
θ
)
sin
θ
d
θ
2
a
n
0
T
0
α
0
2
n
+
1
=
P
n
(
cos
θ
)
sin
θ
d
θ
2
a
n
T
0
1
cos
2
n
+
1
=
P
n
(
x
)
d
x
.
2
a
n
α
P
n
+
1
(
P
n
−
1
(
By using
P
n
(
1
)=
1and
(
2
n
+
1
)
P
n
(
x
)=
x
)
−
x
)
, we arrive at
1
T
0
2
T
0
2
(
a
0
=
d
x
=
1
−
cos
α
)
,
cos
α
T
0
2
a
n
[
a
n
=
P
n
−
1
(
cos
α
)
−
P
n
+
1
(
cos
α
)]
,
n
=
1
,
2
, ··· .
Example 3.
Consider heat conduction in a cylinder
Ω
of radius
a
and height
h
.Its
consists of the cylindrical surface, the upper and the lower circles of
radius
a
. There is no heat source/sink inside the cylinder. The temperature is
f
boundary
∂Ω
(
r
)
(axially symmetric) on the upper circle of
∂Ω
and is zero on the other parts of
∂Ω
.
Find the steady temperature distribution in
Ω
.
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