Cauchy Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 192

The Equation

Substituting ϕ ( x )= 1 into Eq. (5.49), we have

u ( x , t )= u 1 ( x , t )+ u 2 ( x , t )+ u 3 ( x , t ) ,

where

t

e −

u 1 ( x , t )=

2 τ 0

,

τ 0 I 0 b

2 d

x + At

1
2 A

1
2

t

e −

2

u 2 ( x , t )=

2 τ 0

( At ) − ( x − ξ ) ξ ,

x − At

t
2τ 0 I

e −

u 3 ( x , t )=

I 1 b

2 d

x + At

1
8 A

t

t
2τ 0

e −

2

=

b

( At ) − ( x − ξ ) ξ .

2
0

τ

x − At 2 2

( ) − ( − ξ )

At x

By Eq. (5.47), we have

2 1

τ 0

1

t

e − u 2 ( x , t )= − .

Also,

I 1 b

2 d

x + At

1
8 A

t

2

I =

b

( At ) − ( x − ξ ) ξ

2
0

τ x − At 2 2

( At ) − ( x − ξ )

t
τ 0

2 t

τ 0

4 t

τ 0

6

1
8

1
384

1
46080

= + + + ···

t
2

2 t

2

4 t

2

6

1
2!

1
4!

1
6!

= + + + ··· .

τ 0 τ 0 τ 0

i.e.

2 e

τ 0

t 1

t t

e −

τ 0 =

+ = + .

ch

2 τ 0 2

1 I

2

Hence,

2τ 0 ch

1

t
τ 0

e −

t 1

2 −

t t

2τ 0

e − e − u 3 ( x , t )=

τ 0 −

= + .

2 2