Environmental Engineering Reference
In-Depth Information
For
f
(
x
,
y
)=
δ
(
x
−
x
0
,
y
−
y
0
)
,
u
(
x
,
y
)=
v
(
ξ
,
η
;
x
,
y
)
δ
(
ξ
−
x
0
,
η
−
y
0
)
d
ξ
d
η
Δ
M
0
=
v
(
x
0
,
y
0
,
x
,
y
)
.
Therefore, the Riemann function is the solution of PDS (5.10) at
f
(
x
,
y
)=
δ
(
x
−
x
0
,
. The physical meaning of the Riemann function becomes known once the
meanings of
x
,
y
and
u
in PDS (5.10) are available.
y
−
y
0
)
5.1.3 Example
Example.
Use the Riemann method to solve
u
tt
=
a
2
u
xx
+
f
(
x
,
t
)
, −
∞
<
x
<
+
∞
,
0
<
t
,
(5.11)
u
(
x
,
0
)=
ϕ
(
x
)
,
u
t
(
x
,
0
)=
ψ
(
x
)
.
Solution. Step 1. Transform the equation into its standard form.
Let
y
=
at
,
x
=
x
. PDS (5.11) is transformed into
u
xx
−
f
1
x
y
a
a
2
u
yy
=
,
,
f
1
=
−
f
/
, −
∞
<
x
<
+
∞
,
0
<
y
,
(5.12)
u
(
x
,
0
)=
ϕ
(
x
)
,
u
y
(
x
,
0
)=
ψ
(
x
)
/
a
.
Since
Lu
u
yy
,
L
is here a self-conjugate operator.
Step 2. Find the Riemann function.
To find the solution of (5.12) at any point
M
=
Mu
=
u
xx
−
(
x
,
y
)
,
u
(
x
,
y
)
, construct two charac-
teristic curves passing through point
M
:
ξ
+
η
=
x
+
y
,
ξ
−
η
=
x
−
y
,whichin-
terse
ct w
ith the
O
axis at points
P
and
Q
. The domain enclosed by
MP
,
PQ
and
QM
is denoted by
ξ
Δ
M
(Fig. 5.2). By definition (5.8), the Riemann function
M
(
ξ
,
η
)
,
satisfies, for any
M
∈
Δ
M
,
v
(
M
(
x
,
y
))
⎧
⎨
Mv
=
v
ξξ
−
v
ηη
=
0
,
v
(
M
,
M
)=
1
,
MP
=
QM
=
∂
v
,
∂
v
(5.13)
0
0
.
⎩
∂
s
∂
s
It reduces to, by a variable transformation of
x
=
ξ
−
η
,
y
=
ξ
+
η
,
v
x
y
=
0
,
(5.14)
v
|
x
=
x
−
y
=
1
,
v
|
y
=
x
+
y
=
1
.
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