Dual-Phase-Lagging Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 243

where c ( n )
1

and c ( n )
2

are constants. Let

,

e α n t cos e α n t sin β n t β n t

Δ =

e α n t cos e α n t sin ( β n t ) t ( β n t ) t

thus

− −

e α n t sin e α n t sin

β n tf n ( t ) f n ( t ) β n t

c ( n )
1

( t )= d t + c 1 = d t + c 1

β n e 2α n t Δ

t

1
β n

1
β n

e − α n t sin e − α n τ sin

= − ( ) + = − ( τ )

τ .

f n t β n t d t c 1 f n β τ d n

0

Note that

c ( n )
1

e α n t cos

) e α n t cos β n t t c ( n )

1

T n ( t )= ( t ) β n t + ( t

c ( n )
2

e α n t sin

) e α n t sin β n t t c ( n )

2

+ ( t ) β n t + ( t

β n t e α n t cos

) e α n t cos β n t t

1
β n f n (

c ( n )
1

e − α n t sin = − t ) β n t + ( t

c ( n )
2

e α n t sin

) α n e α n t sin β n t . c ( n )

2

e α n t + ( t ) β n t + ( t β n t + β n cos

Thus c ( n )
2

0sothat c ( n )
2

( 0 ) β n = ( 0 )= 0. We then have

e α n t cos

( )

β n tf n t

c ( n )
2

( t )= d t + c 2

Δ

e α n t cos

β n t · f n ( t )

= d t + c 2

β n e 2α n t

t

1
β n

e − α n τ cos

= f n ( τ ) β n τ d τ .

0

Substituting c ( n )
1

and c ( n )
2

( t ) ( t ) into Eq. (6.25) yields

e α n t cos

t

1
β n

e − α n τ sin T n ( t )= − f n ( τ ) β n τ d τ β n t

0

1
β n

e α n t sin

t

e − α n τ cos + f n ( τ ) β n τ d τ β n t

0