Dual-Phase-Lagging Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 287

By Eq. (6.136b)

L t
0

∂Ω

t t

∂
∂

∂
∂

( , , ) | ∂Ω =

τ , τ ,

L u u r u z W f τ d W f τ d W f τ d τ

r z

0 0

∂Ω

L W f τ ,

t

∂
∂

∂
∂

=

r W f τ , z W f τ

d τ = 0 ,

0

so the u ( r , θ , z , t ) in Eq. (6.135) satisfies the boundary conditions of PDS (6.134).

Clearly, the u ( r , θ , z , t ) in Eq. (6.135) satisfies u ( r , θ , z , 0 )= 0. Also,

t

W f τ τ = t ,

∂ W f τ

∂

u t ( r , θ , z , t )= d τ +

t

0

( , θ , , )= ( , θ , , )

which shows that u t r z 0 0 by Eq. (6.136c). Thus the u r z t in Eq. (6.135)

also satisfies the two initial conditions of PDS (6.134).
Since for the u ( r , θ , z , t ) in Eq. (6.135)

t t t

W f τ τ = t =

∂ W f τ

∂

∂ W f τ

∂

u t = ∂
∂

W f τ d τ = d τ + d τ ,

t t t

0 0 0

τ = t

t t

2 W f τ
∂

u tt = ∂
∂

∂
∂

∂

τ + ∂

W f τ
∂ t W f τ d

τ = d

t 2 t t 0 0

t

2 W f τ
∂

∂

= d τ + f ( r , θ , z , t ) ,

t 2

0

t t

0 Δ

= Δ

τ = τ ,

Δ u W f τ d W f τ d

0

t t

0 Δ

∂
∂

= ∂
∂

τ = ∂
∂ t Δ

u

t Δ

W f τ d W f τ d τ

t

0

t t

W f τ τ = t =

∂
∂

∂
∂

=

t Δ

W f τ d τ + Δ

t Δ

W f τ d τ ,

0 0

a substitution of Eq. (6.135) into the equation of PDS (6.134) yields

u t
τ

B 2 ∂
∂

A 2

0 +

u tt − Δ u ( r , θ , z , t ) −

t Δ

u ( r , θ , z , t )

1
τ

d

t

2 W f τ
∂

0 ∂

W f τ
∂ t + ∂

B 2 ∂
∂

A 2 = − Δ W f τ −

t Δ

W f τ τ + f ( r , θ , z , t )

t 2

0

= f ( r , θ , z , t ) .

Thus the u ( r , θ , z , t ) in Eq. (6.135) is indeed the solution of PDS (6.134).