Environmental Engineering Reference
In-Depth Information
Therefore, we obtain, as
ε
→
0
1
1
r
d
S
1
4
r
∂
u
u
∂
∂
1
4
Δ
u
r
u
(
M
0
)=
n
−
−
d
Ω
.
π
∂
n
π
Ω
∂Ω
The third Green formula plays an important role in examining harmonic functions
and in seeking solutions of boundary-value problems of potential equations. Here
we list some of its corollaries.
1. If
u
is a solution of the Poisson equation
Δ
u
=
F
,wehave
1
1
r
d
S
1
4
r
∂
u
u
∂
∂
1
4
F
r
d
u
(
M
0
)=
n
−
−
Ω
.
(7.68)
π
∂
n
π
Ω
∂Ω
2. If
u
is a harmonic function, we have, by Eq. (7.65) and for an internal point
M
0
of
Ω
,
1
1
r
d
S
1
4
r
∂
u
u
∂
∂
u
(
M
0
)=
n
−
,
(7.69)
π
∂
n
∂Ω
which is called the
fundamental integral formula of harmonic functions
.The
value of a harmonic function at any internal point of
Ω
can be thus expressed
by its values and its normal derivatives on the boundary
∂Ω
.
1
3. If
M
0
is outside of
Ω
such that
Δ
r
=
0in
Ω
, the second Green formula leads to
1
1
r
d
S
r
∂
u
u
∂
∂
Δ
u
r
n
−
−
d
Ω
=
0
.
(7.70)
∂
n
∂Ω
Ω
M
0
ε
M
0
ε
If
M
0
is on
∂Ω
,
Ω
refers to the part inside
Ω
.
∂Ω
also refers to the part
M
0
ε
inside
Ω
. Note that the relation between the area element d
S
on
∂Ω
and its
corresponding solid angle d
ω
with respect to the sphere center is
2
d
d
S
=
ε
ω
.
Equations (7.66) and (7.67) thus reduce to
u
r
2
d
S
u
d
S
ε
M
0
ε
=
2
=
u
d
ω
=
u
(
M
ξ
1
)
d
ω
,
M
ξ
1
∈
∂Ω
,
(7.71)
M
0
ε
M
0
ε
M
0
ε
M
0
ε
∂Ω
∂Ω
∂Ω
∂Ω
M
ξ
2
1
r
∂
u
∂
u
d
S
ε
∂
u
M
0
ε
r
d
S
=
ε
2
=
ε
d
ω
,
M
ξ
2
∈
∂Ω
,
(7.72)
∂
∂
r
∂
r
M
0
ε
M
0
ε
M
0
ε
∂Ω
∂Ω
∂Ω
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