Environmental Engineering Reference
In-Depth Information
Fig. 5.2
Domain
Δ
M
Its general solution is
x
,
y
)=
x
)+
y
)
,
(
(
(
v
F
G
where
F
and
G
are differentiable functions of
x
and
y
, respectively. Applying
v
|
x
=
x
−
y
=
1and
v
|
y
=
x
+
y
=
1 yields
y
)=
y
)=
v
(
x
−
y
,
F
(
x
−
y
)+
G
(
1
,
x
,
x
)+
v
(
x
+
y
)=
F
(
G
(
x
+
y
)=
1
,
y
)=
x
)=
which requires
G
(
C
1
(constant),
F
(
C
2
(constant). Therefore
x
,
y
)=
x
,
y
)
∈
Δ
M
.
v
(
C
,
∀
(
Also
v
|
x
=
x
−
y
=
1and
v
|
y
=
x
+
y
=
1
.
M
,
M
∈
Δ
M
.
Thus
v
(
M
)
≡
1
,
Step 3. Solution of PDS (5.12).
Note that
v
≡
1,
b
=
0and
d
η
=
0on
PQ
. Thus the solution of PDS (5.12) is, by
Eq. (5.9),
Q
f
1
d
1
2
[
1
2
1
2
ξ
,
a
u
(
x
,
y
)=
u
(
P
)+
u
(
Q
)]+
u
η
d
ξ
−
ξ
d
η
.
P
Δ
MPQ
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