Dual-Phase-Lagging Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 284

6.6.2 Solution from ϕ ( r , θ , z )

Theorem 1 .Let u ( r , θ , z , t )= W ψ ( r , θ , z , t ) be the solution of

⎨

u t
τ 0 +

B 2 ∂
∂

A 2 u tt = Δ u ( r , θ , z , t )+

t Δ

u ( r , θ , z , t )

Ω × ( 0 , + ∞ ) ,

(6.124)

⎩

L ( u , u r , u z ) | ∂Ω = 0 ,

u ( r , θ , z , 0 )= 0 , u t ( r , θ , z , 0 )= ψ ( r , θ , z ) .

The solution of
⎧
⎨
⎩

u t
τ 0 +

B 2 ∂
∂

A 2

u tt = Δ u ( r , θ , z , t )+

t Δ

u ( r , θ , z , t )

Ω × ( 0 , + ∞ ) ,

(6.125)

L ( u , u r , u z ) | ∂Ω = 0 ,

u ( r , θ , z , 0 )= ϕ ( r , θ , z ) , u t ( r , θ , z , 0 )= 0

is

1 W ϕ (

τ 0 + ∂

B 2 W u ( r , θ , z , t )= r , θ , z , t )+

k ) ϕ (

r , θ , z , t ) ,

2

( k mn + λ

∂ t

where k mn and

2
k are the eigenvalues

λ ( 2 ) and λ ( 3 ) in Section 6.6.1. λ

Proof. By following a similar approach in Section 6.6.1, we have u ( r , θ , z , t ) satis-

fying the equation and the boundary conditions of PDS (6.125)

)= m , n , k e α mnk t u ( r , θ , z , t [( A mnk cos β mnk t + B mnk sin β mnk t ) cos n θ

+( C mnk cos β mnk t + D mnk sin β mnk t ) sin n θ ] J n ( k mn r ) Z k ( z ) . (6.126)

( , θ , , )= ϕ ( , θ , )

Applying the initial condition u r z 0 r z yields

⎧
⎨
⎩

1
M mnk

A mnk = ϕ ( r , θ , z ) J n ( k mn r ) Z k ( z ) r cos n θ d θ d r d z ,

Ω

(6.127)

1
M mnk

C mnk = ϕ ( r , θ , z ) J n ( k mn r ) Z k ( z ) r sin n θ d θ d r d z .

Ω

Applying the initial condition u t ( 0 leads to

α mnk A mnk + β mnk B mnk =

r , θ , z , 0 )=

0 ,

α mnk C mnk + β mnk D mnk = 0 ,