Cauchy Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 179

In order to transform (5.20) into an ordinary differential equation, consider a vari-
able transformation

− ξ − ξ 2 , ξ , τ ∈ Δ M .

2

2

or z 2

2

− ( ξ − ξ )

z = ( t − τ ) =( t − τ ) (5.21)

Thus PDS (5.20) is transformed into
z 2

v (

)+ z ξ ξ − z ττ v ( z 2

τ

c 2 v

ξ −

z z )+ ( z )= 0 ,

(5.22)

v ( 0 )= 1 .

Note that, by Eq. (5.21)

zz ξ = ξ − ξ ,

zz τ = − ( t − τ ) ,

which lead to, by subtracting the square of the former from the square of the latter,

z 2 z 2

ξ − τ = −

1 . (5.23)

Also, from Eq. (5.21)

z 2 z 2

τ +

zz ττ = 1 ,

ξ +

zz ξ ξ = − 1 ,

which yield, by subtracting the former from the latter,

z z ξ ξ − z ττ = −

z 2 z 2

ξ − τ +

2 . (5.24)

Substituting Eq. (5.23) into Eq. (5.24) yields

1
z .

z ξ ξ − z ττ = − (5.25)

Fig. 5.3 Domain Δ

M