Dual-Phase-Lagging Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 235

The solution of the well-posed PDS
⎧
⎨

u t
τ 0 +

B 2 ∂
∂

A 2 u tt = Δ u +

t Δ

u ,

Ω × ( 0 , + ∞ ) ,

(6.4)

⎩

L ( u , u n ) | ∂Ω = 0 ,

u ( M , 0 )= ϕ ( M ) , u t ( M , 0 )= 0 .

is

1 W ϕ (

τ 0 + ∂

u = M , t )+ W ϕ 1 ( M , t ) , (6.5)

∂ t

B 2

where ϕ 1 = − Δϕ .

Proof. By the definition of W ϕ ( M , t ) and W ϕ 1 ( M , t ) ,wehave

⎧
⎨

2 W ϕ
∂ τ 0 ∂

W ϕ
∂

+ ∂

1

B 2 ∂
∂

A 2

−

W ϕ − W ϕ =

,

Δ

t Δ

0 (6.6a)

t t 2

∂Ω =

L W ϕ , ∂

W ϕ
∂

0 , (6.6b)

⎩

n

t = 0 = ϕ (

W ϕ t = 0 =

∂ W ϕ

∂

0 , M ) . (6.6c)

t

⎧
⎨

2 W ϕ 1
∂ τ 0 ∂

W ϕ 1
∂ t + ∂

1

B 2 ∂
∂

A 2 − Δ W ϕ 1 −

t Δ

W ϕ 1 = 0 , (6.7a)

t 2

∂Ω =

L W ϕ 1 , ∂

W ϕ 1
∂

0 , (6.7b)

⎩

n

t = 0 = ϕ

W ϕ 1 t = 0 =

∂ W ϕ 1

∂

0 , ( M ) . (6.7c)

1

t

Satisfaction of Equation

Substituting Eq. (6.5) into the equation of PDS (6.4) yields

1
τ

W ϕ +

W ϕ 1

u t
τ

B 2 ∂
∂

1
τ

∂
∂ 0 + ∂

A 2

0 +

u tt − Δ u −

t Δ

u =

t ∂ t

0

1 W ϕ +

W ϕ 1

1 W ϕ +

W ϕ 1

2

+ ∂

τ 0 + ∂ τ 0 + ∂

A 2 − Δ

∂ t 2 ∂ t ∂ t

1 W ϕ +

W ϕ 1

B 2 ∂
∂ τ 0 + ∂

−

t Δ ∂ t