Dual-Phase-Lagging Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 281

6.6 Mixed Problems in a Cylindrical Domain

Boundary conditions of all the three kinds for mixed problems in a cylinder become
separable with respect to the spatial variables in a cylindrical coordinate system. In
this section we apply separation of variables to seek solutions to mixed problems in
a cylindrical coordinate system,
⎧
⎨

u t
τ 0 +

B 2 ∂
∂

A 2 u tt = Δ u ( r , θ , z , t )+

t Δ

u ( r , θ , z , t )

+ f ( r , θ , z , t ) , Ω × ( 0 , + ∞ ) ,

(6.114)

⎩

L ( u , u r , u z ) | ∂Ω = 0 ,

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

where the Ω stands for a cylindrical domain: 0 < r < a , 0 < z < H ,the ∂Ω is the

boundary of Ω . If all combinations of boundary conditions of all three kinds are

, there exist 27 combinations. We will also examine the relation
among solutions from

considered on ∂Ω

ϕ ( r , θ , z ) , ψ ( r , θ , z ) and f ( r , θ , z , t ) , respectively.

6.6.1 Solution from ψ ( r , θ , z )

The solution due to ψ ( r , θ , z ) satisfies

⎧
⎨

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 , + ∞ ) ,

(6.115)

⎩

L ( u , u r , u z ) | ∂Ω = 0 ,

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

Assume u = T ( t ) v ( r , θ , z ) . Substituting this into the equation of PDS (6.115) yields,

λ ( 1 ) as the separation constant, with

T +

τ 0 T

1

B 2 T = Δ

v
v = − λ ( 1 ) . A 2 T

+

Thus we arrive at

1

τ 0 + λ ( 1 ) B 2 T + λ ( 1 ) A 2 T

T +

= 0 (6.116)