Cauchy Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 193

Finally,

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

t
τ 0
2 +

t
τ 0
2 +

e − e −

1
2 −

1
2 −

t
2τ 0

t
2τ 0

e − e −

= + = 1 ,

so it satisfies the equation of PDS (5.48). By this verification, we find another
integral

⎡

I 0 b

2

x + At

1
2

⎣

2

( At ) − ( x − ξ )

τ

x − At 0

2 ⎤

I 1 b

(5.50)

t

2

⎦ d

+

0 b

( At ) − ( x − ξ ) ξ

2 2

4 τ ( At ) − ( x − ξ )

2 A e 1

t
2τ 0

= − .

5.3.3 Verify the Solution for f ( x , t )= 1 , u ( x , 0 )= 0 and u t ( x , 0 )= 0

Verify that the solution of

⎧
⎨

u t
τ 0 +

A 2 u xx

u tt = + 1 ,

(5.51)

⎩

u ( x , 0 )= 0 , u t ( x , 0 )= 0

is

I 0 b A 2

2 d

t x + A ( t − τ )

1
2 A

t − τ
2

e −

2

u ( x , t )=

τ 0 d

τ ( t − τ ) − ( x − ξ ) ξ . (5.52)

0 x − A ( t − τ )

τ = By the variable transformation t − τ , Eq. (5.52) becomes

I 0 b

2 d

t

τ x + A τ
x − A τ

e − τ

1
2 A

2

u ( x , t )=

2τ 0 d

( A τ ) − ( x − ξ ) ξ .

0

It reduces to, using Eq. (5.47),

0 τ 0 1

τ 0 d

t

0 e −

1

e − τ

t
τ 0

τ = τ 0 t

2

u ( x , t )= − + τ − . (5.53)

It is straightforward to show that u ( x , t ) in Eq. (5.53) satisfies PDS (5.51).