Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 134

First find the Green function G ( x , s ) that satisfies

⎨

k

( , )

d
d x

d G x s

( x ) = δ ( x − s ) ,

d x

(3.44)

⎩

G ( 0 , s )= G ( l , s )= 0 .

k

d
d x

d G ( x , s )

The general solution of ( x ) = 0is

d x

G

= ( )+ ( ) , , .

c 1 u 1 x c 2 u 2 x c 1 c 2 are constants

where

d x

u 1 ( x )= 1 , u 2 ( x )=

) . k ( x

Let

(

a 1 + b 1 ) u 1 +( a 2 + b 2 ) u 2 , x ≤ s ,

G ( x , s )= (3.45)

( a 1 − b 1 ) u 1 +( a 2 − b 2 ) u 2 , x ≥ s .

Applying the property of G and d G
d x

at x = s ,

s + , s − , d G ( s ) d G ( s ) 1

s + , s − ,

G ( s ) − G ( s )= 0 , − =

d x d x k ( s )

yields

u 2 ( s ) − 1

b 1 =

) ,

b 2 =

) .

2 u 2 ( 2 u 2 (

s ) k ( s s ) k ( s

( , )= ( , )=

Applying the boundary conditions G 0 s G l s 0 thus leads to

a 2 = −

2 b 1 − b 2 [ u 2 ( 0 )+ u 2 ( l )]

a 1 = − b 1 − ( a 2 + b 2 ) u 2 ( 0 ) , .

u 2 ( 0 ) − u 2 ( l )

With the Green function G ( x , s ) , we transform PDS (3.42) into an eigenvalue prob-

lem of the integral equation

l

X ( x )= − λ G ( x , s ) X ( s ) d s . (3.46)

0