Mixed Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 146

τ = t −

A 2 t

2 W f τ
∂

t t

1
τ

∂ W f τ

∂

∂

τ + ∂

W f τ
∂

= d τ + d

0 Δ

W f τ d τ

t t 2 t

0 0 0

1 d

2 W f τ
∂

t

τ 0 ∂

W f τ
∂ t + ∂

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

t 2

0

which indicates that u 3 satisfies the equation of PDS (4.4).
Also,

∂Ω = ∂Ω

L u 3 , ∂ L

u 3
∂

t

∂
∂

t

W f τ d τ , W f τ d τ

n n

0 0

L W f τ , ∂

∂Ω

t

W f τ
∂

= d τ = 0 ,

n

0

which shows that the u 3 satisfies the boundary conditions of PDS (4.4).
Finally,

t = 0 =

W f τ τ = t t = 0 =

t

0and ∂

u 3
∂

∂ W f τ

∂

u 3 ( M , 0 )= d τ + 0 .

t t

0

t

Therefore, the u 3 also satisfies the initial conditions so u 3 = W f τ ( M , t − τ ) d τ

0

0.
3. Since PDS (4.4) is linear, the principle of superposition is valid. Applying this
principle to PDS (4.4) concludes that u

is indeed the solution of PDS (4.4) at ϕ = ψ =

= u 1 + u 2 + u 3 is the solution of PDS (4.4).

Remark 1 . The solution of
⎧
⎨

a 2 u t + τ 0 u tt = Δ u + f ( M , t ) , Ω × ( 0 , + ∞ ) ,

∂Ω =

L u , ∂

u

(4.6)

0 ,

⎩

∂ n

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

can still be written as

1
τ

W ϕ (

t

0 + ∂

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

∂ t

0

=

W ψ (

, )

ϕ =

=

Here u 2 M t is the solution of PDS (4.6) at f 0;

1 W ϕ (

τ 0 + ∂

u 1 = M , t ) is the solution of PDS (4.6) at ψ = f = 0;

∂ t

t

u 3 = W f τ ( M , t − τ ) d τ is the solution of PDS (4.6) at ϕ = ψ = 0. The f τ is defined

0

1
τ

by f τ = f ( M , τ ) .

0