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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