Dual-Phase-Lagging Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 303

Substituting (6.172) into (6.171) and applying (6.172) yield the solution of (6.169),

N
k = 0

N
n = 0 C n a n x n − k G k ( t )

1

t

2 A e −

u =

2 τ 0

N

n = 0 a n x n G 0 ( t )+

N
k = 1 n = 0 a n ( x n

N

) ( k )
k !

1 1

t
2τ 0

t
2τ 0

2 A e − 2 A e −

=

G k (

)

t

1

τ 0

P ( k )
N

N
k = 1

1 ( x )

t t

e −

2 A e −

= τ 0 P N ( x ) − +

2 τ 0

G k ( t )

k !

m
k = 1

1

τ 0

P ( 2 k )
N

1 ( x )

t t

e −

2 A e −

= τ 0 P N ( x ) − +

2 τ 0

! G 2 k

. (6.174)

( 2 k )

2. When N is odd such that N 1, the solution of PDS (6.169) can be formed

by the sum of that in Eq. (6.174) and the solution due to the last term of P N (

= 2 m +

x )=

N
n = 0 a n x n ,i.e. a 2 m + 1 x 2 m + 1 . The latter can be obtained by Eq. (6.168),

I 0 b

2 a 2 m + 1 ξ

τ 0 x + At
x

1

t

2 A e −

2 m + 1 d

2

u =

2

( At ) − ( x − ξ ) ξ

− At

u 2 a 2 m + 1

k = 0 C 2 m + 1 u k x 2 m + 1 − k d u

I 0 b

τ 0 + At
−

2 m + 1

1

t

2 A e −

2

=

2

( At ) −

At

a 2 m + 1 x 2 m + 1 ( k )
k !

2 m 1

k = 1

+

1 1

t t

2 A e − 2 A e −

τ 0 a 2 m + 1 x 2 m + 1 G 0 (

=

2

t )+

2 τ 0

G k ( t )

a 2 m + 1 x 2 m + 1 ( 2 k )
(

= τ 0 a 2 m + 1 x 2 m + 1 1

τ 0

m
k = 1

1

t t

e −

2 A e −

− +

2 τ 0

G 2 k ( t ) .

)

2 k !

(6.175)

Thus the solution of PDS (6.169) is, by adding Eqs. (6.174) and (6.175),

[ N / 2 ]
k = 1

1

τ 0

P ( 2 k )
N

1 ( x )

t t

2τ 0

e −

2 A e −

u = τ 0 P N ( x ) − +

! G 2 k (

t ) , (6.176)

( 2 k )

where [ N / 2 ] stands for the maximum positive integer not larger than N / 2. Equa-

tion (6.176) shows that the solution is a sum of [ N / 2 ]+ 1 terms. All the terms are

in the form of separation of variables.
3. Denote the solution (6.176) of PDS (6.169) by u = W P N ( x , t ) . The solution of

u t / τ 0 +

A 2 u xx + R 1

u tt = P N ( x ) , × ( 0 , + ∞ ) ,

(6.177)

u ( x , 0 )= P m ( x ) , u t ( x , 0 )= P l ( x )