Dual-Phase-Lagging Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 282

and

Δ

)+ λ ( 1 ) v

v ( r , θ , z ( r , θ , z )= 0 , ( r , θ , z ) ∈ Ω ,

(6.117)

L ( v , v r , v z ) | ∂Ω = 0 , v ( r , θ + 2 π , z )= v ( r , θ , z ) .

λ ( 2 ) as the sepa- Let v = V ( r , θ ) Z ( z ) . Substituting this into Eq. (6.117) yields, with

ration constant,

Z ( Δ V ( r , θ ) z )

) − λ ( 1 ) = − λ ( 2 ) .

, θ ) = − V ( r Z ( z

Thus we have

, θ )+ λ ( 2 ) V

Δ V ( r ( r , θ )= 0 , 0 < r < a ,

(6.118)

L ( V , V r ) | r = a = 0 , V ( r , θ + 2 π )= V ( r , θ ) .

and

Z (

)+ λ ( 3 ) Z , λ ( 3 ) = λ ( 1 ) − λ ( 2 ) , z ( z )= 0

(6.119)

L ( Z , Z z ) | z = 0 = 0 , L ( Z , Z z ) | z = H = 0 .

The solution of the former is available in Section 4.3.2

V mn ( r , θ )= J n ( k mn r )( a mn cos n θ + b mn sin n θ ) ,

( n m are determined by Eq. (6.97),

where a mn and b mn are constants, the μ

a 0 2 μ ( n ) λ ( 2 ) = k mn =

/ , m = 1 , 2 , ··· , n = 0 , 1 , 2 , ··· . (6.120)

m

There are a total of nine combinations of boundary conditions in Eq. (6.119). Let
λ ( 3 ) = λ

2
k

stand for the eigenvalues and eigenfunctions of Eq. (6.119),
which are available in Table 2.1. Substituting

and Z k ( z )

λ ( 1 ) = λ ( 2 ) + λ ( 3 ) = k mn + λ

2
k

into

( )

Eq. (6.116) leads to the T t -equation

1
τ 0 +

k B 2 T mnk (

k mn + λ k mn + λ

k A 2 T mnk (

T mnk ( t )+ t )+ t )= 0 . (6.121)

Its characteristic roots read

1
τ 0 +(

1
τ 0 +(

B 2 B 2 2 k mn + λ

k

k mn + λ

2

k mn + λ

2

A 2 − ) ±

k )

− 4 (

k )

r 1 , 2 =

2

= α mnk ± β mnk i . (6.122)

e α mnk t Thus T mnk ( t )= ( c mnk cos β mnk t + d mnk sin β mnk t ) .