Cauchy Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 230

5.8.4 Discussion

Unit Analysis

c
A L , e −

2τ 0

T − 1
LT − 1 =

1

t

In Eq. (5.120), [ b ]= = = 1, [ I 1 ]= 1. Therefore, the unit

of the first term in the right-hand side of Eq. (5.120)

1

A 2 1

T 2
L 2

1

T · T ·

L 2 = [ ψ ][ d S ]= · = Θ .

t

The unit of the second term

1
A

] dr

1

= [ b ] [ ψ ][ d S

2

r 2

( At ) −

T
L ·

1
L ·

1

L · T ·

L 2 = · L = Θ .

Therefore, the unit of Eq. (5.120) is correct.

Initial Conditions

Note that

1
At

M )

ψ ( d S = 4 π At · ψ ( ¯ At ) and lim ψ ( ¯ At )= ψ ( M ) .

t → + 0

S At

The u ( M , t ) in Eq. (5.120) thus satisfies the initial condition u ( M , 0 )= 0. Note also

that

I 1 ( x = x 1

2 .

lim

t → + 0

Hence

⎡
⎣

1
2

1

t

τ 0 e −

u t ( M , t )= −

2 τ 0

4 π At · ψ ( ¯ At )

4 π A

⎛
⎝

⎞
⎠

⎤
⎦

bI 1 b ( r 2

At

At )

2

−

1
2

M ) d r

(

+ ψ ( dS

At )

2

− r 2

− At

S M
r