Environmental Engineering Reference
In-Depth Information
Consider Neumann problems of Laplace equations
⎧
⎨
Δ
u
=
0
,
(
x
,
y
)
∈
D
,
∂
D
=
(7.179)
∂
u
⎩
f
(
x
,
y
)
.
∂
n
By the necessary condition for the existence of solutions,
f
(
x
,
y
)
d
s
=
0
,
∂
D
∂
D
=
v
∂τ
∂
where
s
is the arclength. Since
f
(
x
,
y
)
, the monotropic continuous function
defined on
∂
D
(
x
,
y
)
v
(
x
,
y
)=
f
(
x
,
y
)
d
s
+
v
(
x
0
,
y
0
)
(7.180)
(
x
0
,
y
0
)
can be determined for fixed point
A
(
x
0
,
y
o
)
∈
∂
D
. Thus
v
(
x
,
y
)
must satisfy
=
,
(
,
)
∈
,
Δ
v
0
x
y
D
(7.181)
v
|
∂
D
=
F
(
x
,
y
)
.
(
x
,
y
)
where
F
(
x
,
y
)=
f
(
x
,
y
)
d
s
+
v
(
x
0
,
y
0
)
. Once the solution of Dirichlet prob-
(
x
0
,
y
0
)
lem (7.181) is available, we may obtain the solution
u
(
x
,
y
)
of PDS (7.179) by
Eq. (7.178). The
u
(
x
,
y
)
so obtained is unique up to an arbitrary constant.
Remark 4
. Green functions for Dirichlet problems of two-dimensional potential
equations can be found by using a mirror image method if the domain has some
symmetry. If the domain is lacking such symmetry, we can first apply a conformal
transformation to transform the domain into a symmetrical one. Let be a simply-
connected plane domain with a smooth boundary. Suppose that
w
i
y
is a complex-valued variable) is a conformal transformation that is able to transform
D
into a unit circle
=
w
(
z
)
(
z
=
x
+
|
w
| <
1. Let
z
0
=
x
0
+
i
y
0
be an internal point of
D
. The conformal
transformation
w
(
z
)
−
w
(
z
0
)
w
=
w
(
z
,
z
0
)=
1
−
w
(
z
)
w
(
z
0
)
will map
D
into the unit circle
|
w
| <
1and
z
0
into the center of
|
w
| <
1. The solution
of
−
Δ
G
(
x
,
y
;
x
0
,
y
0
)=
δ
(
x
−
x
0
,
y
−
y
0
)
,
(
x
,
y
)
∈
D
,
G
|
∂
D
=
0
,
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