Cauchy Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 218

Remark 1. An odd continuation should be applied if the boundary condition is
u ( 0 , t )= 0.

Remark 2. In Eq. (5.97),

b

2

L − 1

[ b ]= , ( At )

2

− ( r − ρ ) = 1 , [ I 0 ]= 1 ,

1
Ar

TL − 2 T − 1 [ u ]= [ ρ ][ ψ ( ρ )] [ d ρ ]= · L · Θ · L = Θ

Remark 3. It is straightforward to show that the u in Eq. (5.97) satisfied u ( r , 0 )= 0.

Note that I 0 ( 0 )= 1. Also

⎧
⎨
⎩ −

t

I 0 b

2

τ 0 r + At
r − At

τ 0 e −

1
2 Ar

1
2

2

u t ( r , t )= ( At )

2

− ( r − ρ ) ρψ ( ρ ) d ρ

t

τ 0 r + At
r − At

e −

∂ I 0

∂

2

+ ρψ ( ρ ) d ρ +( r + At ) ψ ( r + At ) A

t

A +( r − At ) ψ ( r − At ) .

Thus u t ( r , 0 )= ψ ( r ) .

Remark 4. An examination of the right-hand side of Eq. (5.97) shows that the u ( r , t )

in Eq. (5.97) is indeed an even function of r . By L'Hôpital's rule, we have

τ 0 lim
r → 0

r + At

t

1 I 1 ( y )

2 A e −

2

lim

r → 0 u

( r , t )= ( ρ − r ) ρψ ( ρ ) d ρ

y r − At

+( r + At ) ψ ( r + At ) − ( r − At ) ψ ( r − At )

τ 0 At
−

t

1 I 1 ( y 1 )

y 1 2 A e −

2

=

2

ψ ( ρ ) ρ + ψ (

) ,

ρ d 2 At At

At

b b

where y = ( At )

2

− ( r − ρ )

2 and y 1

= ( At )

2

− ρ

2 . By the mean value theorem

of integrals,
At

At

I 1 ( y 1 )

y 1

d ρ

2 2

ρ ψ ( ρ ) d ρ = I 1 ( y 1 ) ρ ¯ ψ ( ρ ) ¯

b

( At )

2

− ρ

2

− At − At

¯

2

¯

I 1 ( y 1 ) ρ ψ ( ρ ) π

= ,

b

b

2 2 ,

where y 1 = ( At ) − ρ ¯ − At < ρ < ¯ At . Thus lim

r 0 u

( r , t ) does exist and r = 0isa

→

removable discontinuous point.