Environmental Engineering Reference
In-Depth Information
Fig. 7.3
Symmetric point
M
1
(
r
1
cos
θ
0
,
r
1
θ
0
)
of
M
0
(
r
0
cos
θ
0
,
r
0
θ
0
)
For any point
P
on the circle
r
=
R
,wehave
Δ
OM
0
P
Δ
OPM
1
so that
r
M
0
P
r
0
R
=
R
r
1
=
r
M
1
P
.
Therefore
ln
R
r
0
1
r
M
0
P
=
R
r
0
1
r
M
1
P
1
2
1
r
M
0
P
=
1
2
1
r
M
1
P
or
ln
,
π
π
which yields
ln
R
r
0
1
2
1
r
M
1
M
g
(
M
,
M
0
)=
.
π
ln
ln
R
r
0
.
1
2
1
r
M
0
P
−
1
r
M
1
M
Finally, the Green function is
G
(
M
,
M
0
)=
π
Also
r
=
R
=
∂
r
=
R
=
−
R
2
r
0
−
∂
G
G
∂
1
r
0
.
∂
n
r
2
π
R
R
2
−
2
Rr
0
cos
(
θ
−
θ
0
)+
Thus the solution of PDS (7.117) in a polar coordinate system is
2π
R
2
r
0
1
2
−
u
(
r
0
,
θ
0
)=
f
(
θ
)
d
θ
.
(7.119)
r
0
−
π
R
2
+
2
Rr
0
cos
(
θ
−
θ
0
)
0
This is the same as that obtained in Section 7.2 by the method of separation of vari-
ables and in Section 7.4 by the relation between harmonic functions and analytical
functions.
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