Environmental Engineering Reference
In-Depth Information
Note that the equation in PDS (7.59) is a second-order homogeneous Euler equation.
Its general solution can be readily obtained
C
1
r
+
=
,
v
C
2
where
C
1
and
C
2
are constants. Applying the boundary conditions
v
(
R
1
)=
v
0
and
v
(
R
2
)=
0 yields
R
1
R
2
R
2
−
R
1
R
2
−
−
C
1
=
R
1
v
0
,
C
2
=
R
1
v
0
.
Thus the solution of (7.59) is
1
r
−
v
0
.
R
1
R
2
R
2
1
R
1
v
(
r
)=
−
R
1
Example 3
. Consider the distribution of electric potential
u
in a sphere of radius 1.
The potential distribution on its boundary surface is
u
cos
2
|
r
=
1
=
θ
. Find the poten-
tial distribution in the sphere.
Solution.
By the given conditions, the electric potential
u
cannot depend on
ϕ
in a
spherical coordinate system
(
r
,
θ
,
ϕ
)(
0
≤
ϕ
≤
2
π
)
. Thus the
u
must satisfy
r
2
∂
sin
⎧
⎨
r
2
∂
1
u
1
r
2
sin
∂
∂θ
θ
∂
u
∂θ
+
=
,
<
θ
<
π
,
<
<
,
0
0
0
r
1
∂
r
∂
r
θ
⎩
cos
2
u
|
r
=
1
=
θ
.
(7.60)
=
(
)
Θ
(
θ
)
Assume
u
R
r
. Substituting it into the Laplace equation in PDS (7.60)
yields
r
2
R
+
2
rR
=
−
Θ
+(
θ
)
Θ
cot
=
λ
,
R
Θ
where
λ
is the separation constant. Let
λ
=
n
(
n
+
1
)
. Thus we have the Euler and
the Legendre equations
r
2
R
+
2
rR
−
n
(
n
+
1
)
R
=
0
,
Θ
+(
θ
)
Θ
+
cot
n
(
n
+
1
)
Θ
=
0
.
They have the bounded solutions
C
n
r
n
in
r
≤
1and
P
n
(
cos
θ
)(
n
=
0
,
1
,
2
, ···
)
in
[
, respectively. Here the
C
n
are constants. Thus we obtain the solution of the
Laplace equation in PDS (7.60)
0
,
π
]
∞
n
=
0
C
n
r
n
P
n
(
cos θ
)
.
u
=
cos
2
Applying the boundary condition
u
|
r
=
1
=
θ
yields
∞
n
=
0
C
n
P
n
(
x
)
,
d
n
dx
n
(
1
2
n
n
!
x
2
x
2
n
=
P
n
(
x
)=
−
1
)
,
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