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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