Mixed Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 145

1
τ

1
τ

2 W ϕ
∂

2 W ϕ
∂

1
τ

0 ∂

W ϕ
∂

+ ∂ + ∂

∂

0 ∂

W ϕ
∂

+ ∂ A 2 A 2 = − Δ W ϕ − Δ W ϕ = 0 ,

t t 2 t t t 2

0

which indicates that u 1 satisfies the equation of PDS (4.4) at f = 0. Also

L u 1

∂Ω =

L 1

W ϕ , 1 W ϕ ∂Ω

, ∂

u 1
∂ τ 0 + ∂

∂
∂ τ 0 + ∂

n ∂ t n ∂ t

∂Ω + ∂Ω

L W ϕ , ∂

∂
∂ t L W ϕ , ∂

1
τ

W ϕ
∂

W ϕ
∂

=

n n

0

τ 0 L W ϕ ,

∂Ω + ∂ L W ϕ , ∂ ∂Ω

1 ∂

∂

W ϕ
∂

=

n W ϕ

= 0 .

∂ t n

This shows that u 1 satisfies the boundary conditions of PDS (4.4).
Finally,

1 W ϕ t = 0 = t = 0 + ∂ t = 0 = ϕ (

W ϕ
∂ τ 0 + ∂

1
τ 0 W ϕ

( , )= ) .

u 1 M 0 M

∂ t t

Also,

t = 0 = ∂ 1 W ϕ t = 0 = 1 t = 0

2 W ϕ
∂

∂ u 1

∂ τ 0 + ∂ τ 0 ∂

W ϕ
∂ t + ∂

t 2

t ∂ t ∂ t

W ϕ t = 0 = A 2

= Δ 0 .

1 W ϕ (

τ 0 + ∂

= , )

Therefore, the u 1 M t is the solution of PDS (4.4)

∂ t

= ψ =

at f 0.

2. Since W f τ (

, − τ )

M t satisfies

⎧
⎨

2 W f τ
∂ τ 0 ∂

1 W f τ

∂ t + ∂

A 2

= Δ W f τ , Ω , 0 < τ < t < + ∞ ,

t 2

L W f τ , ∂

∂Ω =

W f τ
∂

0 ,

n

⎩

t = τ =

W f τ t = τ =

∂
∂

0 ,

t W f τ

f ( M , τ ) ,

therefore

2 u 3
∂

τ 0 ∂

1 u 3

∂ t + ∂

1
τ 0

∂
∂

t

A 2

t 2 −

Δ u 3 = W f τ ( M , t − τ ) d τ

t 0

t 2

2

+ ∂

t t

A 2 W f τ ( M , t − τ ) d τ − Δ W f τ ( M , t − τ ) d τ

∂

0 0