Mixed Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 144

Theorem. Let u 2 = W ψ ( M , t ) denote the solution of

⎧
⎨

A 2

/ τ + = ,

Ω × (

, + ∞ ) ,

u t u tt Δ u 0 0

L u

∂Ω =

, ∂

u

0 , (4.3)

⎩ ∂ n

u ( M , 0 )= 0 , u t ( M , 0 )= ψ ( M )

The solution of

⎧
⎨

A 2

u t / τ 0 + u tt = Δ u + f ( M , t ) , Ω × ( 0 , + ∞ ) ,

L u

∂Ω =

, ∂

u

0 , (4.4)

⎩ ∂ n

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

can be written as

1 W ϕ (

τ 0 + ∂

t

u = u 1 + u 2 + u 3 = M , t )+ W ψ ( M , t )+ W f τ ( M , t − τ ) d τ ,

∂ t 0

(4.5)

1
τ

W ϕ (

0 + ∂

where f τ = f ( M , τ ) .Here u 1 = M , t ) is the solution of (4.4) at

∂ t

t

f = ψ = 0. u 3 = W f τ ( M , t − τ ) d τ is the solution of (4.4) at ϕ = ψ = 0.

0

Proof.
1. As W ϕ (

, )

M t satisfies

⎧
⎨

2 W ϕ
∂ τ 0 ∂

1 W ϕ

∂ t + ∂

A 2 = Δ W ϕ , Ω × ( 0 , + ∞ ) ,

t 2

L W ϕ , ∂

∂Ω =

W ϕ
∂

0

⎩ n

t = 0 = ϕ (

W ϕ t = 0 =

∂ W ϕ

∂

, ) .

0 M

t

Hence

2 u 1
∂

τ 0 ∂

1 u 1

∂ t + ∂

A 2

t 2 −

Δ u 1

1 W ϕ + ∂

t 2 1

W ϕ − 1 W ϕ

2

1
τ 0

∂
∂ τ 0 + ∂ τ 0 + ∂ τ 0 + ∂

A 2

= Δ

t ∂ t ∂ ∂ t ∂ t