Environmental Engineering Reference
In-Depth Information
t
sin
f n
1
β n
1
2
e α n ( t τ )
=
( τ +
)+
( τ
)
( τ )
β
t
sin
β
t
d
τ
n
n
0
sin
f n ( τ )
t
1
β n
e α n ( t τ ) 1
2
+
β n (
t
+ τ )+
sin
β n (
t
τ )
d
τ
0
t
1
β n
e α n ( t τ ) sin
=
(
τ )
( τ )
β
t
f n
d
τ
n
0
l
t
2
e α n ( t τ ) sin
=
d
ξ
β n (
t
τ )
f
( ξ , τ )
d
τ .
(6.26)
l
β n
0
0
A substitution of Eq. (6.26) into Eq. (6.23) yields the solution of PDS (6.22),
l
t
+
n = 1
2
e α n ( t τ ) sin n
π
x
sin n
πξ
l
u
(
x
,
t
)=
d
ξ
sin
β n (
t
τ )
f
( ξ , τ )
d
τ .
l
β n
l
0
0
Let
+
n = 1
2
l
1
β n
e α n ( t τ ) sin n
π
x
sin n
πξ
l
G
(
x
, ξ ,
t
τ )=
sin
β n (
t
τ ) .
(6.27)
l
The solution of PDS (6.22) becomes
l
t
u
(
x
,
t
)=
d
ξ
G
(
x
, ξ ,
t
τ )
f
( ξ , τ )
d
τ .
(6.28)
0
0
The G
in Eq. (6.27) can be called the Green function of one-dimensional
dual-phase-lagging heat-conduction equations . It is clearly boundary-condition de-
pendent. When f
(
x
, ξ ,
t
τ )
(
x
,
t
)= δ (
x
x 0 ,
t
t 0 )
, Eq. (6.28) reduces to
u
=
G
(
x
,
x 0 ,
t
t 0 ) ,
so that the Green function G
(
x
, ξ ,
t
τ )
is the temperature distribution exclusively
from the source term
.
The solution of PDS (6.14) is, by the principle of superposition, the sum of those
in Eqs. (6.18), (6.21) and (6.28).
δ (
x
ξ ,
t
τ )
Solve PDS (6.19) and verify its solution
u
is available in Eq. (6.18). By Theorem 1 in Section 6.1, we have the
solution of PDS (6.19)
=
W ψ (
x
,
t
)
1
τ
W ϕ (
0 +
=
u
x
,
t
)+
W B 2 Δϕ (
x
,
t
) ,
(6.29)
t
 
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