Environmental Engineering Reference
In-Depth Information
The solution of
u t
τ 0 +
B 2
A 2
u tt =
Δ
u
(
r
, θ ,
t
)+
t Δ
u
(
r
, θ ,
t
)+
f
(
r
, θ ,
t
) ,
0
<
r
<
a
,
0
< θ <
2
π ,
0
<
t
,
(6.109)
L
(
u
,
u r ) | r = a =
0
,
u
(
r
, θ ,
0
)=
0
,
u t (
r
, θ ,
0
)=
0
is
t
u
(
r
, θ ,
t
)=
W f τ (
r
, θ ,
t
τ )
d
τ ,
(6.110)
0
where f τ =
f
(
r
, θ , τ )
. Therefore, Theorem 2 in Section 6.1 is also valid in polar
coordinate systems.
Proof. By the definition of W ψ (
r
, θ ,
t
)
,the W f τ =
W f τ (
r
, θ ,
t
τ )
satisfies
2 W f τ
τ 0
1
W f τ
t +
B 2
A 2
=
Δ
W f τ +
t Δ
W f τ ,
(6.111a)
t 2
L W f τ ,
r = a =
W f τ
0
,
(6.111b)
r
t = τ =
W f τ t = τ =
0
,
t W f τ
f
(
r
, θ , τ ) .
(6.111c)
By Eq. (6.111b), we have
L t
0
r = a
t
L
(
u
,
u r ) | r = a =
W f τ d
τ ,
W f τ d
τ
r
0
r = a
L W f τ ,
t
=
r W f τ
d
τ =
0
,
0
so that the u
(
r
, θ ,
t
)
in Eq. (6.110) satisfies the boundary conditions of PDS (6.109).
Clearly, the u
(
r
, θ ,
t
)
in Eq. (6.110) satisfies u
(
r
, θ ,
0
)=
0. Also,
t
W f τ τ = t ,
W f τ
u t (
r
, θ ,
t
)=
d
τ +
t
0
whichiszeroat t
=
0, by Eq. (6.111c). Therefore the u
(
r
, θ ,
t
)
in Eq. (6.110) also
satisfies the two initial conditions of PDS (6.109).
 
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