Environmental Engineering Reference
In-Depth Information
Example 2.
Find the Green function in the upper half space
Ω
:
z
>
0 and solve the
boundary-value problem
Δ
u
=
F
(
M
)
,
M
∈
Ω
,
(7.116)
u
|
z
=
0
=
f
(
M
)
.
Solution.
For any point
M
0
(
x
0
,
y
0
,
z
)
∈
Ω
, its symmetric point with respect to the
boundary
z
=
0is
M
1
(
x
0
,
y
0
,−
z
0
)
. The Green function can be readily obtained as
1
4
1
G
(
M
,
M
0
)=
π
2
2
2
(
x
−
x
0
)
+(
y
−
y
0
)
+(
z
−
z
0
)
1
−
(
,
x
−
x
0
)
2
+(
y
−
y
0
)
2
+(
z
+
z
0
)
2
where
M
(
x
,
y
,
z
)
∈
Ω
. Clearly, the
G
satisfies
G
|
z
=
0
=
0. Also,
z
=
0
=
−
∂
z
=
0
=
−
∂
G
G
z
0
2
1
/
2
.
3
∂
n
∂
z
π
2
2
z
0
]
[(
x
−
x
0
)
+(
y
−
y
0
)
+
Thus the solution of PDS (7.116) is
)
∂
G
(
)=
−
(
,
,
−
(
,
)
(
,
,
)
u
M
0
f
x
y
0
n
d
S
G
M
M
0
F
x
y
z
d
Ω
∂
z
=
0
z
>
0
+
∞
+
∞
z
0
2
f
(
x
,
y
,
0
)
=
d
x
d
y
(
z
0
π
3
/
2
−
∞
−
∞
x
−
x
0
)
2
+(
y
−
y
0
)
2
+
+
∞
+
∞
+
∞
1
4
1
−
(
π
x
−
x
0
)
2
+(
y
−
y
0
)
2
+(
z
−
z
0
)
2
0
−
∞
−
∞
F
1
(
−
(
x
,
y
,
z
)
d
x
d
y
d
z
.
x
−
x
0
)
2
+(
y
−
y
0
)
2
+(
z
+
z
0
)
2
Example 3
. Find the Green function in the quarter space
Ω
:
x
>
0
,
y
>
0
,−
∞
<
z
<
+
∞
.
Solution.
Consider a point electric charge of capacity
ε
at
M
0
∈
Ω
. The electric
1
4π
r
M
0
M
.Here
potential at
M
(
x
,
y
,
z
)
∈
Ω
is thus
r
M
0
M
=
(
x
−
x
0
)
2
+(
y
−
y
0
)
2
+(
z
−
z
0
)
2
.
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