Environmental Engineering Reference
In-Depth Information
The counterpart of PDS (5.99) in classical heat-conduction equations is
u
t
a
2
=
Δ
u
(
r
,
t
)
,
0
<
r
<
+
∞
,
0
<
t
,
(5.102)
u
r
(
0
,
t
)=
0
,
u
(
r
,
0
)=
1
.
After an even continuation and a function transformation
v
(
r
,
t
)=
ru
(
r
,
t
)
,it
reduces to
v
t
=
a
2
Δ
v
(
r
,
t
)
, −
∞
<
r
<
+
∞
,
0
<
t
,
(5.103)
v
(
r
,
0
)=
r
.
Its solution (also the solution of PDS (5.102)) is available in Chapter 3 as
+
∞
+
∞
2
2
4
a
2
t
e
−
(
r
−
ρ
)
e
−
ξ
1
2
a
√
π
1
2
a
√
π
u
(
r
,
t
)=
4
a
2
t
ρ
d
ρ
=
(
ξ
−
r
)
d
ξ
tr
tr
−
∞
−
∞
+
∞
+
∞
2
4
a
2
t
d
e
−
ξ
1
2
a
√
π
1
√
π
2
e
−
η
=
ξ
=
d
η
=
1
.
t
−
∞
−
∞
Therefore, this property is also preserved by the hyperbolic heat-conduction
equation.
2. By Eq. (5.98), the solution of
⎧
⎨
u
t
τ
0
+
)
τ
0
A
2
u
tt
=
Δ
u
(
r
,
t
)+
f
(
r
,
t
,
0
<
r
<
+
∞
,
0
<
t
,
u
r
(
0
,
t
)=
0
,
(5.104)
⎩
u
(
r
,
0
)=
0
,
u
t
(
r
,
0
)=
0
is
t
u
=
W
f
τ
(
r
,
t
−
τ
)
d
τ
0
I
0
b
2
t
r
+
A
(
t
−
τ
)
t
−
τ
2
1
2
A
e
−
2
=
τ
0
d
τ
(
At
)
−
(
r
−
ρ
)
ρ
f
(
ρ
,
τ
)
d
ρ
.
τ
0
r
0
r
−
A
(
t
−
τ
)
(5.105)
When
f
(
r
,
t
)=
δ
(
r
−
r
0
,
t
−
t
0
)
,wehave
τ
0
r
I
0
b
A
2
2
e
−
r
0
2
A
t
−
t
0
2
2
u
(
r
,
t
)=
(
t
−
t
0
)
−
(
r
−
r
0
)
τ
0
,
(5.106)
t
0
)
τ
0
,thetem-
which shows that under the effect of a point source
δ
(
r
−
r
0
,
t
−
perature is higher inside the spherical surface
r
=
r
0
than that outside the surface,
and is relatively lower on the surface
r
=
r
0
.Furthermore,
ρ
1
2
c
lim
r
→
0
u
(
r
,
t
)=
∞
,
lim
r
u
(
r
,
t
)=
0
,
lim
r
u
(
r
,
t
)=
0
.
(5.107)
k
τ
→
∞
→
r
0
t
→
t
0
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