Dual-Phase-Lagging Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 247

can be obtained by the superposition of Eqs. (6.18) and (6.21)
⎧
⎨

+ ∞

n = 1 ( A n cos β n t + B n sin β n t ) e α n t sin n π x

u ( x , t )= ,

l

2

l 0 ϕ (

sin n

π x

A n = x ) d x ,

(6.33)

⎩ l

l
0 ψ (

l
0 ϕ (

2

sin n

π x 2 α

sin n

π x n

B n = x ) d x − x ) d x .

l β n l l β n l

To prove that the u in Eq. (6.33) is indeed a solution of PDS (6.32), we must

show that we can obtain u tt , u xx and u xxt from Eq. (6.33) by taking derivatives term
by term. Thus we must prove that the series u

( x , t )

( x , t ) , u tt ( x , t ) , u xx ( x , t ) and u xxt ( x , t ) are

all uniformly convergent in ( 0 , l ) × ( 0 , T ) ,where T is an arbitrary positive constant

and t .

Let the general term of series u

∈ [ 0 , T ]

( x , t ) (Eq. (6.33)) be u n ( x , t ) ,i.e.

e α n t sin n

π x

l ,

u n ( x , t )=( A n cos β n t + B n sin β n t )

thus

n 2 2

∂ l

e α n t sin n

π x

x 2 u n (

x , t )= − ( A n cos β n t + B n sin β n t ) ,

∂ l

t

2

∂

)= ∂
∂

t 2 u n

( x , t ( − A n β

n sin

β

n t

+ B n β n cos β

n t

+ A n α

n cos

β

n t

∂

e α n t sin n

π x

+ B n α

n sin

β

n t

)

l

A n

β n cos

= ∂
∂

+

α B n β n t n

t

β n t e α n t sin n

π x

+( B n α n − A n β n ) sin

l

−

A n

β n sin

B n

β n cos

= + +

β n −

α β B n β β n t α A n β β n t

n n n n n

2
n

+( A n α + B n α n β n ) cos β n t

β n t e α n t sin n

π x

2
n

+( B n α − A n α n β n ) sin

l

A n

β n n cos

2
n

= α − β + 2 B n α β β n t n n

A n

n sin +( − 1 ) α β + B n β n β β n t n n

n t e α n t sin n

π x

2
n +( B n α − A n α β ) sin β ,

n n

l

n 2 A n α n +

B n β n cos

3

∂ l

x 2 u n (

x , t )= − β n t

∂ t ∂

e α n t sin n

π x

( B n α − A n β ) sin β n t ] . n n

l