Environmental Engineering Reference
In-Depth Information
Substituting Eqs. (5.23) and (5.25) into (5.22) leads to
1
z v (
v (
c 2 v
z
)+
z
)
(
z
)=
0
,
v
(
0
)=
1
.
ξ ,
Therefore, v depends on
τ
,
ξ
and t only through z . By another variable transfor-
i cz i
=
1 , it becomes
mation
η =
1
η
v ( η )+
v ( η )+
v
( η )=
0
,
v
(
0
)=
1
,
where the equation is the Bessel equation of order zero. Its solution is (see Ap-
pendix A)
J 0 i c
2
I 0 c
2
v ξ ,
ξ , τ =
2
2
( ξ ξ )
( ξ ξ )
t ;
(
t
τ )
=
(
t
τ )
,
where J 0 is the Bessel equation of order zero of the first kind. I 0 is the modified
Bessel function of order zero of the first kind.
Find solutions of PDS (5.18) and (5.16)
The solution of PDS ( 5.1 8) follows from Eq. (5.9) in Section 5. 1, by noting that
f
ξ τ 0 )
( ξ ,
0and u τ ( ξ ,
(
x
,
t
)=
0, d
τ =
0on PQ , u
0
)=
0
)= ψ (
a
,
I 0 c t 2
2
ξ + t
ψ a
ξ τ 0 d
1
2
ξ .
U
( ξ ,
t
)=
( ξ ξ )
ξ t
The solution of PDS (5.17) is
t
0 U
e
u
( ξ ,
t
)=
( ξ ,
t
) .
ξ = τ
/
Since
0 x
a , the solution of PDS (5.16) is
u
=
W ψ (
x
,
t
)
a ξ 2
τ 0
a
τ 0 x
a
d
(5.26)
ξ
τ 0
1
2
x
+
t
t
0
e
c
ψ
ξ .
=
I 0
t 2
τ 0
a
x
t
Finally, the solution of PDS (5.15) is, by the solution structure theorem,
1
W ϕ (
t
τ 0 +
u
=
x
,
t
)+
W ψ (
x
,
t
)+
W f τ (
x
,
t
τ )
d
τ
t
0
, τ ) τ 0 .
where f τ =
f
(
x
Search WWH ::




Custom Search