Dual-Phase-Lagging Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 290

Assume u = T ( t ) v ( r , θ , ϕ ) . Substituting it into the equation of PDS (6.141) yields,

k 2 as the separation constant,

with − λ = −

1

T +

τ 0 T

B 2 T = Δ

v ( r , θ , ϕ )

k 2

, θ , ϕ ) = − λ = −

.

A 2 T

+ v ( r

Thus we arrive at

1
τ 0 + λ

B 2 T ( T ( A 2 T

t )+ t )+ λ ( t )= 0 (6.142)

and

k 2 v

( , θ , ϕ )+ ( , θ , ϕ )= , < < , < θ < π ,

Δ v r r 0 0 r a 0

(6.143)

L ( v , v r ) | r = a = 0 .

The solution of the latter is available in Section 2.6.2

P n ( v mnl ( r , θ , ϕ )=( a mnl cos m ϕ + b mnl sin m ϕ ) cos θ ) j n ( k nl r ) , m ≤ n ,

a 2

n

2

1

+

k nl =

where constants a mnl and b mnl are not all zero, λ = μ , n =

l

n +

2

1

0 , 1 , 2 , ··· , l = 1 , 2 , ··· and the μ are available in Section 2.6.2. Substituting

l

a 2

n +

2

1

k nl =

λ = μ into Eq. (6.142) yields

l

1
τ 0 +(

2 T nl (

T
nl

2 T nl (

( t )+ k nl B ) t )+( k nl A ) t )= 0 .

Its characteristic roots read

1
τ 0 +(

1
τ 0 +(

2 2 2 2

− k nl B ) ± k nl B ) − 4 ( k nl A )

r 1 , 2 = (6.144)

2

= α nl ± β nl i .

Thus

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