Cauchy Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 204

Example 2. Solve

Lu

R 2

= 0 , × ( 0 , + ∞ ) ,

(5.74)

u ( x , y , 0 )= 1 , u t ( x , y , 0 )= 0 .

Solution. By Eq. (5.69) and the result in Example 1, we obtain

ch A

1
2

( At )

2

− ( x − ξ )

2

− ( y − η )

2

1

τ 0 + ∂

t

e −

(

u ( x , y , t )=

2 τ 0

d ξ d η

2 π A ∂ t

At )

2

− ( x − ξ )

2

− ( y − η )

2

D At

e − ct c

1

2 c e ct

e − ct = + ∂

∂

= − 1 .

t

This is also the same as its one-dimensional counterpart. It is also physically
grounded by noting that the u

( , , ) ( , , )=

x y t comes only from u x y 0 1.

From this special case, we obtain

ch A ( At )

2

− ( x − ξ )

2

− ( y − η )

2

∂
∂

d ξ d η

t

2 2 2

( At ) − ( x − ξ ) − ( y − η )

D At

ch A (

At )

2

− ( x − ξ )

2

− ( y − η )

2

1
2

t
2τ 0

= −

η .

2 π A e d ξ d

τ 0

2 2 2

( At ) − ( x − ξ ) − ( y − η )

D At

Example 3. Solve

Lu

R 2

= 1 , × ( 0 , + ∞ ) ,

(5.75)

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

Solution. By Eq. (5.70) and the result in Example 1, we obtain

t

t − τ

2τ 0

e −

u ( x , y , t )=

0

⎡ ⎤

ch A A 2

⎣

2 2 2

⎦

( t − τ ) − ( x − ξ ) − ( y − η )

1

· A 2 d ξ d η d τ

2 π A

2 2 2

( t − τ ) − ( x − ξ ) − ( y − η )

D A ( t − τ )

t

e − c ( t − τ ) e c ( t − τ ) − e − c ( t − τ ) d

1
2 c

= τ

0

= τ 0 t
0

1

τ 0 d

0 e −

1

t − τ t

τ 0

e −

2

− τ = τ 0 t + τ − . (5.76)

This is also the same as its one-dimensional counterpart (Eq. (5.53) in Section 5.3).