Environmental Engineering Reference
In-Depth Information
Fig. 5.1
Domain
Δ
M
0
d
s
√
2whered
s
is the differential
On
M
0
M
1
,
x
−
y
=
x
0
−
y
0
so d
y
=
d
x
=
−
of arclength
s
. Thus
M
0
M
1
−
Q
d
x
+
P
d
y
=
M
0
M
1
[(
uv
)
y
−
(
2
v
y
+
bv
)
u
]
d
x
+[(
uv
)
−
(
2
v
x
−
av
)
u
]
d
y
x
(5.5)
2
∂
v
b
a
√
2
uv
d
s
−
=
d
(
uv
)
−
s
uds
+
∂
M
0
M
1
2
∂
√
2
v
u
d
s
−
v
b
a
=
uv
|
M
1
−
uv
|
M
0
−
s
−
.
∂
M
0
M
1
Similarly,
2
∂
v
u
d
s
v
a
b
√
2
+
M
2
M
0
−
Q
d
x
+
P
d
y
=
uv
|
M
2
−
uv
|
M
0
+
s
+
.
(5.6)
∂
M
2
M
0
Substituting Eqs. (5.5) and (5.6) into Eq. (5.3) and solving for
(
uv
)
|
M
0
yields
∂
2
√
2
v
u
d
s
)
|
M
2
1
2
v
a
+
b
(
uv
)
|
M
0
=
(
uv
)
|
M
1
+(
uv
+
s
+
∂
M
2
M
0
∂
2
√
2
v
u
d
s
v
b
−
a
1
2
−
s
−
+
M
1
M
2
−
Qdx
+
Pdy
(5.7)
∂
M
0
M
1
1
2
−
(
vLu
−
uMv
)
d
σ
,
Δ
M
0
where
u
|
M
0
=
u
(
x
0
,
y
0
)
is what we attempt to find.
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