Cauchy Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 181

Remark 1. Note that I 0 (

x )= I 1 ( x ) . The solution of PDS (5.15) at f = ψ = 0 is thus

ϕ x at √ τ 0 + ϕ x at √ τ 0

1 W ϕ (

− +

τ 0 + ∂

t
2τ 0

e − u = x , t )=

∂ t 2

√ τ 0
a

1
2

a

√ τ 0 ξ d

c 2 t
y

1
2

x + t

ξ

+

τ 0 I 0 (

y )+ I 1 ( y ) ϕ , (5.27)

√ τ 0
a

x − t

√ τ 0
a

− ξ 2 t 2 where y = c − x . Also, the solution of PDS (5.15) at ψ = ϕ = 0

reads

t

u = W f τ ( x , t − τ ) d τ

0

c

− ξ 2

√ τ 0
a

2

I 0 ( t − τ ) − x

√ τ 0
a

f a

d

t

ξ
√ τ 0 , τ

x + t

1
2

t − τ

2τ 0 d

e −

ξ . = τ √ τ a x − t

τ 0

0

(5.28)

[ ξ ]=[ √ τ 0 x [ ξ ]= Since / a ]= T ,wehave T , [ y ]= 1and [ I 0 ]= 1. Therefore,

ξ ]= Θ T − 1 Eq. (5.26): [ u ]=[ I 0 ][ ψ ][ d · T = Θ .

c 2 t
y

ξ T − 1 Eq. (5.27): [ u ]= I 1 ϕ d = Θ T = Θ (The unit of all the other terms is

also clear.)

d

ξ

I 0
τ 0 f d

T − 1 T − 1 T Eq. (5.28): [ u ]= τ = T · · Θ = Θ .

Remark 2. Rewrite PDS (5.15) into
⎧
⎨

u t
τ

f ( x , t )

A 2 u xx +

0 +

u tt = , − ∞ < x < + ∞ , 0 < t ,

τ (5.29) 0

⎩

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

a √ τ 0 [ LT − 1 is the velocity of thermal waves. For a variable

where A = A ]=

√ τ 0
a ξ

ξ = transformation ,

√ τ 0
a

1
A d

ξ = ξ = ξ , d d

c t 2 b

√ τ 0
a

− ξ 2 2 a √ τ 0 .

2 2

y = − x = ( At ) − ( x − ξ ) , b = 1 /