Environmental Engineering Reference
In-Depth Information
Consider a function
v
(
x
,
y
)
such that
⎨
Mv
=
0
,
v
|
M
0
=
1
,
2
√
2
v
M
0
M
1
=
2
√
2
v
M
2
M
0
=
∂
∂
(5.8)
v
b
−
a
v
a
+
b
⎩
s
−
0
,
s
+
0
.
∂
∂
Clearly, such a function
v
depends on point
M
0
, and thus is denoted by
v
(
x
,
y
;
x
0
,
y
0
)
(
,
)
(
,
,
)
(
,
)
or
v
defined by Eq. (5.8) is called the
Riemann function of operator
L
. For the Riemann function
v
defined by Eq. (5.8),
Eq. (5.7) yields
M
M
0
. The function
v
x
y
;
x
0
y
0
or
v
M
M
0
⎧
⎨
⎩
)
|
M
2
1
2
u
(
x
0
,
y
0
)=
(
uv
)
|
M
1
+(
uv
+
M
1
M
2
−
[
−
(
uv
)
y
+(
2
v
y
+
bv
)
u
]
d
x
⎬
⎭
.
(5.9)
+[(
uv
)
x
−
(
2
v
x
−
av
)
u
]
d
y
−
v
(
x
,
y
;
x
0
,
y
0
)
f
(
x
,
y
)
d
σ
Δ
M
0
Remark 1.
Once the Riemann function
v
is available, the right-hand side
of Eq. (5.9) becomes known. The task of solving PDS (5.4) reduces to that of finding
the Riemann function from Eq. (5.8). Such a method of solving PDS is called the
Riemann method
.
(
x
,
y
;
x
0
,
y
0
)
M
1
M
2
of curve
c
,
Remark 2.
On
u
x
=
u
s
cos
(
x
,
s
)+
u
n
cos
(
x
,
n
)
d
x
d
s
2
u
x
d
x
d
y
d
s
)] =
ϕ
(
=
d
s
+
u
n
[
−
sin
(
x
,
s
x
)
−
ψ
(
x
)
1
2
ϕ
(
f
(
f
(
x
)
−
ψ
(
x
)
x
)
+[
x
)]
=
.
1
+[
f
(
x
)]
2
Similarly,
)
1
u
y
=
ϕ
(
f
(
x
)
x
)+
ψ
(
x
+[
f
(
x
)]
2
.
f
(
2
1
+[
x
)]
Remark 3.
By using Eq. (5.9), the solution of the PDS
Lu
=
−
(
,
)
f
x
y
(5.10)
|
c
=
|
c
=
u
u
n
0
(
,
)=
(
ξ
,
η
,
)
(
ξ
,
)
η
.
reads
u
x
y
v
;
x
y
f
y
d
ξ
d
Δ
M
0
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