Environmental Engineering Reference
In-Depth Information
also be determined by applying the initial condition
u
t
|
t
=
0
=
Ψ
(
r
,
θ
,
z
)
. Finally,
⎧
⎨
u
=
W
Ψ
(
r
,
θ
,
z
,
t
)
b
(
1
)
J
n
∞
∑
sin
μ
l
z
t
2τ
0
e
−
b
(
2
)
=
θ
+
(
)
,
mnl
cos
n
mnl
sin
n
θ
k
mn
r
H
sin
γ
mnl
t
n
=
0
,
m
,
l
=
1
r
sin
μ
l
z
1
b
(
1
)
m
0
l
=
Ψ
(
r
,
θ
,
z
)
J
0
(
k
m
0
r
)
H
d
z
d
r
d
θ
,
2
πγ
m
0
l
M
m
0
l
Ω
⎩
r
sin
μ
l
z
1
πγ
mnl
M
mnl
b
(
1
)
mnl
=
Ψ
(
r
,
θ
,
z
)
J
n
(
k
mn
r
)
H
cos
n
θ
d
z
d
r
d
θ
,
Ω
r
sin
μ
l
z
1
πγ
mnl
M
mnl
b
(
2
)
mnl
=
Ψ
(
r
,
θ
,
z
)
J
n
(
k
mn
r
)
H
sin
n
θ
d
z
d
r
d
θ
.
Ω
Thus the solution of PDS (4.60) is, by the solution structure theorem,
1
W
Φ
(
t
τ
0
+
∂
u
=
r
,
θ
,
z
,
t
)+
W
Ψ
(
r
,
θ
,
z
,
t
)+
W
F
τ
(
r
,
θ
,
z
,
t
−
τ
)
d
τ
,
∂
t
0
,
τ
)
τ
0
,
M
mnl
=
where
F
τ
=
F
(
r
,
θ
,
z
M
mn
M
l
.
μ
(
n
m
a
0
2
λ
l
=
μ
l
H
2
n
2
Remark.
The eigenvalues
λ
mn
=
,
λ
n
=
and
cor-
respond to the eigenfunctions
J
n
μ
(
n
m
r
a
0
,
a
n
cos
n
0
and sin
μ
l
z
H
, respectively. Similar to in a cube, they come from the triple
application of separation of variables. Here, we also have the
T
θ
a
n
+
b
n
=
θ
+
b
n
sin
n
(
t
)
-equation [and
Eq. (4.59)]. For the case of a cube,
λ
=
λ
m
+
λ
n
+
λ
l
. For the case of a cylinder,
(
n
m
a
0
2
+
μ
l
H
2
. The former can be obtained ei-
ther by separation of variables three times or by first expanding the solution into
a series and then substituting it into the equation. However, the latter can be proven
only by separation of variables three times.
however,
λ
=
λ
+
λ
l
=
μ
mn
4.4.3 Spherical Domain
As described in Section 2.6.2, for mixed problems in a sphere of radius
a
0
, bound-
ary conditions are separable only in a spherical coordinate system. The differ-
ence from Section 2.6.2 lies in the
T
(
t
)
-equation. After separating variables by
u
=
T
(
t
)
U
(
r
,
θ
,
ϕ
)
, we have, with
−
λ
standing for the separation constant,
τ
0
T
+
T
+
λ
a
2
T
=
0
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