Environmental Engineering Reference
In-Depth Information
5.4.5 Special Cases
For conciseness, we define
2
2
2
1
τ
∂
∂
t
+
∂
Δ
=
∂
x
2
+
∂
A
2
L
=
t
2
−
Δ
,
y
2
.
∂
∂
∂
0
Example 1.
Solve
Lu
R
2
=
0
,
×
(
0
,
+
∞
)
,
(5.73)
u
(
x
,
y
,
0
)=
0
,
u
t
(
x
,
y
,
0
)=
1
.
Solution.
By Eq. (5.67), we obtain
ch
A
2τ
0
2
2
2
1
(
At
)
−
(
x
−
ξ
)
−
(
y
−
η
)
t
A
e
−
u
(
x
,
y
,
t
)=
(
d
ξ
d
η
2
π
At
)
2
−
(
x
−
ξ
)
2
−
(
y
−
η
)
2
D
At
ch
A
2τ
0
2π
0
At
2
r
2
1
(
At
)
−
t
A
e
−
=
d
θ
r
d
r
2
π
(
At
)
2
−
r
2
0
c
e
−
ct
(
At
)
A
√
v
d
c
A
√
v
A
√
v
2
2
1
ch
c
1
c
e
−
ct
sh
c
(
At
)
=
=
0
0
e
−
ct
=
τ
0
1
τ
0
2
c
e
−
ct
e
ct
1
t
e
−
=
−
−
,
1
where
c
τ
0
. Therefore, the temperature is independent of spatial coordinates.
This is the same as in the one-dimensional case. Also, lim
t
=
2
u
=
τ
0
,sothat
τ
0
is, in
→
∞
its value, equal to the temperature of any point at
t
→
+
∞
. This can serve as one
method of measuring
τ
0
.
From this special case, we obtain the integral
ch
A
2
2
2
(
At
)
−
(
x
−
ξ
)
−
(
y
−
η
)
d
ξ
d
η
(
At
)
2
−
(
x
−
ξ
)
2
−
(
y
−
η
)
2
D
At
ch
A
2π
At
2
r
2
(
At
)
−
=
d
θ
r
d
r
(
At
)
2
−
r
2
0
0
τ
0
1
τ
0
e
t
t
2τ
0
e
−
=
2
π
A
−
.
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