Environmental Engineering Reference
In-Depth Information
Since
t
2
r
t
2
r
2
2
=
∂
∂
2
d
left-hand side
d
τ
u
d
S
=
4
π
u
(
τ
,
t
)
τ
τ
,
∂
∂
0
0
S
M
0
r
a
2
S
M
0
r
∂
u
a
2
r
2
∂
u
(
r
,
t
)
right-hand side
=
r
d
S
=
4
π
,
∂
∂
r
we arrive at
t
2
r
2
(
,
)
∂
a
2
r
2
∂
u
r
t
2
d
u
(
τ
,
t
)
τ
τ
=
.
∂
∂
r
0
Taking the derivative with respect to
r
yields
r
2
∂
2
ur
2
(
)
∂
∂
∂
u
a
2
=
,
∂
t
2
r
∂
r
i.e.
2
v
2
v
∂
a
2
∂
t
2
=
r
2
,
(2.86)
∂
∂
where
v
(
r
,
t
)=
r u
(
r
,
t
)
. The general solution of Eq. (2.86) reads, by Sect. 2.7.2,
v
(
r
,
t
)=
f
(
r
+
at
)+
g
(
r
−
at
)
,
where
f
and
g
are two differentiable functions. Applying
v
(
0
,
t
)=
0 leads to
g
(
−
at
)=
−
f
(
at
)
so that
v
(
r
,
t
)=
r u
=
f
(
r
+
at
)
−
f
(
at
−
r
)
.
(2.87)
By taking derivatives of Eq. (2.87) with respect to
r
and
t
, respectively, we obtain
f
(
f
(
u
+
r u
r
=
r
+
at
)+
at
−
r
)
,
(2.88)
1
a
∂
(
r u
)
f
(
f
(
=
r
+
at
)
−
at
−
r
)
.
(2.89)
∂
t
Adding Eqs. (2.88) and (2.89) leads to
)=
∂
(
r u
)
1
a
∂
(
r u
)
2
f
(
r
+
at
r
+
,
∂
∂
t
=
=
and for
t
0and
r
at
0
,
∂
(
r
=
at
0
,
t
=
0
.
r u
)
1
a
∂
(
r u
)
2
f
(
)=
+
at
0
∂
r
∂
t
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