Environmental Engineering Reference
In-Depth Information
whichtendto2
π
u
(
M
0
)
and zero, respectively, as
ε
→
0. because as
ε
→
0,
M
ξ
→
∂
M
0
.
∂
u
u
d
ω
→
2
π
,
u
(
M
ξ
)
→
u
(
M
0
)
,
∂
r
∂
r
M
0
ε
∂Ω
Equation (7.65) becomes
u
∂
∂
1
r
d
S
1
2
1
r
∂
u
1
2
Δ
u
r
u
(
M
0
)=
−
−
−
d
Ω
.
(7.73)
π
n
∂
n
π
∂Ω
Ω
Equations (7.68), (7.69) and (7.73) can be summarized as
⎧
⎨
4
π
u
(
M
0
)
,
M
0
∈
Ω
,
u
1
r
d
S
)
∂
∂
1
r
∂
u
(
M
)
Δ
u
r
−
(
M
−
−
d
Ω
=
2
π
u
(
M
0
)
,
M
0
∈
∂Ω
,
n
∂
n
⎩
¯
∂Ω
Ω
0
,
M
0
∈
Ω
.
(7.74)
=
(
where
r
=
M
0
M
x
−
x
0
)
2
+(
y
−
y
0
)
2
+(
z
−
z
0
)
2
.If
u
satisfies
Δ
u
=
0, in par-
ticular, the triple integral on the left-hand side of Eq. (7.74) vanishes.
The counterpart of Eq. (7.74) in a two-dimensional case reads
⎧
⎨
2
π
u
(
M
0
)
,
M
0
∈
D
,
u
ln
1
r
ln
1
r
∂
d
S
∂
∂
u
Δ
u
r
(
)
,
∈
∂
,
−
(
M
)
−
−
d
σ
=
π
u
M
0
M
0
D
⎩
n
∂
n
D
D
0
,
M
0
∈
.
(7.75)
∂
D
This can be obtained by following a
simi
lar approach to that used in developing
Eq. (7.74), simply replacing
Ω
,
∂Ω
,
M
0
M
and the solid angle 4
π
of a sphere by a
plane
domain
D
,
the
piecewise
smooth
boundary
curve
2
2
and the central angle 2
∂
of a circle, and using the
two-dimensional fundamental solution ln
r
.When
u
satisfies
D
,
r
=
(
x
−
x
0
)
+(
y
−
y
0
)
π
=
Δ
u
0, the double
integral in the left-hand side of Eq. (7.75) reduces to zero.
Remark.
For a fixed point
M
0
R
3
and a variable point
M
R
3
,
(
x
0
,
y
0
,
z
0
)
∈
(
x
,
y
,
z
)
∈
1
r
satisfies the harmonic equation in
R
3
except at the point
M
0
the function
v
(
M
)=
so
1
r
=
Δ
0
,
r
=
0
.
Thus
1
r
d
1
r
d
1
r
1
r
3
r
Δ
Ω
=
Δ
Ω
=
Δ
·
n
d
S
=
·
n
d
S
r
≤
ε
r
=
ε
r
=
ε
R
3
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