Cauchy Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 225

F − 1

¯

Note that [ ψ ( ω )] = ψ ( M ) ,

sin t ( c 2

F − 1 ⎡

⎤

A ω )

2

−

⎣ ⎦

(

A ω )

2

− c 2

sin t c 2

(

ω )

2

−

A

1

e iω · r d

= ( ω 1 d ω 2 d ω 3

3

( 2 π )

A ω )

2

− c 2

R 3

⎛

2 ⎞
⎠ e − iωβ d

A

i r e iω r cos θ

+ ∞ At π

1 2 A 1

⎝ c

=

t 2

− −

I 0 β d ω

( 2 π )

2

0 − At

0

⎡ ⎛

2 ⎞
⎠ cos

⎤

A

+ ∞ At

1 1

Ar

⎣ ⎝ c ⎦ sin

= ω I 0 t 2 − ωβ d β ω r d ω

( 2 π )

2

− At

0

r + At c t 2 d

+ ∞

1

− (

r − ρ )

2

= I 0 ω sin ωρ d ρ ω

2 Ar

A 2

4 π

0 r − At

= v 1 ( M , t ) , (5.114)

where we have used Eq. (5.111). Therefore, by the convolution theorem

v ( M , t )= v 1 ( M , t ) ∗ ψ ( M ) .

Finally, the solution of
⎧
⎨

u t
τ 0 +

A 2 R 3

= + ( , , , ) , × ( , + ∞ ) ,

u tt Δ u f x y z t 0

(5.115)

⎩

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

is, by the solution structure theorem,

1 W ϕ (

t

τ 0 + ∂

u ( x , y , z , t )= M , t )+ W ψ ( M , t )+ W f τ ( M , t − τ ) d τ

∂ t

0

⎡
⎣ e −

1
τ

τ 0

0 + ∂

t

=

2

v 1 ( ξ , η , ζ , t ) ϕ ( x − ξ , y − η , z − ζ ) d ξ d η d ζ

∂ t

R 3