Cauchy Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 227

infinitesimal solid angle. Thus

1
4

r 2 d M , d S = ω and v ( r , t )= v ( t ) d ω .

π

S r

Integrating the equation in PDS (5.113) over V r yields

A 2
V r

c 2
V r

2 v

∂

t 2 d V

= Δ v d V + v d V .

∂

V r

Since

⎛
⎝

⎞
⎠

t 2 r

2 v 2

∂

= ∂

t 2 d V

v d S d ρ ,

∂ ∂ 0

V r

S M
ρ

A 2
V r

A 2
S r

∂ v

Δ vdV =

n d S

∂

A 2 r 2
S r

∂ v

A 2 r 2 1
4

∂ v

=

n d

ω = 4 π

r d

ω

∂ π ∂

S r

A 2 r 2 ∂

v

= 4 π ,

∂ r

⎛
⎝

⎞
⎠

c 2
V r

c 2 r
0

v d V = v d S d ρ ,

S M
ρ

we arrive at

⎛ ⎞ ⎛ ⎞

A 2 r 2 ∂

t 2 r

c 2 r
0

2

∂ v

⎝ ⎠ ⎝ ⎠

v d S d ρ = 4 π + v d S d ρ .

∂ ∂ r

0

S M
ρ

S M
ρ

Taking the derivative with respect to r yields

r 2 ∂

2 v

∂

A 2 1

r 2 ∂

v

c 2 v

t 2 =

+ .

∂ ∂ r ∂ r