Environmental Engineering Reference
In-Depth Information
Example 5.
Find the harmonic function in a circle of radius
R
and with its boundary
value
k
0
cos
R
.Here
k
0
is a given constant.
Solution.
The required harmonic function is the solution of
⎧
⎨
θ
on the circle
r
=
2
u
2
u
∂θ
,
θ
)=
∂
1
r
∂
u
r
2
∂
1
Δ
u
(
r
r
2
+
r
+
2
,
0
<
r
<
R
,
∂
∂
(7.120)
⎩
u
(
R
,
θ
)=
k
0
cos
θ
.
By the Poisson formula (7.119), the solution of PDS (7.120) is
2π
R
2
r
2
1
2
−
θ
θ
u
(
r
,
θ
)=
k
0
cos
d
R
2
r
2
(
θ
−
θ
)
π
+
−
2
Rr
cos
0
1
d
2π
k
=
1
r
k
2
∞
cos
k
θ
−
θ
k
0
2
θ
θ
=
cos
+
π
R
0
∞
d
2π
k
=
1
r
R
k
cos
k
θ
k
0
π
θ
θ
cos
k
θ
sin
k
θ
=
θ
+
cos
sin
k
0
2π
k
0
π
r
R
cos
k
0
R
r
cos
cos
2
θ
d
θ
=
=
θ
θ
,
0
wherewehaveusedRemark1inExample 1 in Section 7.2.
Remark 1
. We can also obtain the solution of PDS (7.120) by using the method
of undetermined coefficients. It is straightforward to show that
u
,
where
A
is a constant, satisfies the Laplace equation in PDS (7.120). Applying the
boundary condition
u
(
r
,
θ
)=
Ar
cos
θ
k
0
(
r
,
θ
)=
AR
cos
θ
=
k
0
cos
θ
yields
A
=
R
. The solution of
PDS (7.120) is thus
k
0
R
r
cos
u
(
r
,
θ
)=
θ
.
Remark 2
. In a Cartesian coordinate system, PDS (7.120) reads
⎧
⎨
2
u
2
u
∂
x
2
+
∂
x
2
y
2
R
2
y
2
=
0
,
+
<
,
∂
∂
⎩
k
0
x
u
|
x
2
=
R
.
y
2
R
2
+
=
k
R
.
Consider the solution
u
=
Ax
. Applying the boundary condition yields
A
=
Therefore the solution is
k
0
R
x
k
0
R
r
cos
u
=
=
θ
.
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