Potential Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 354

Thus the PDS (7.42) is transformed into
⎧
⎨

Δ v = 0 , 0 < x < a , 0 < y < b ,

v x ( 0 , y )= v ( a , y )= 0 , (7.43)

⎩

v ( x , 0 )= f ( x ) , v ( x , b )= g ( x ) .

where

q
2 (

1
2 (

x 2 a 2 x 2 a 2 f ( x )= − − ) , g ( x )= − − )( pb + q ) . (7.44)

By Table 2.1 , we can expand v ( x , y ) by

∞
k = 0 Y k ( y ) cos λ k x ,

v =

) π
2 a . Substituting this into the Laplace equation in PDS (7.43)
yields the ordinary differential equation Y k (

λ k = (

2 k + 1

where

y ) , whose general solution can be readily

determined. Finally, we obtain v so that it satisfies the Laplace equation and the

boundary conditions with respect to x in PDS (7.43).

( x , y )

∞
k = 0 ( a k e λ k y

b k e − λ k y

v = + ) cos λ k x , (7.45)

where constants a k and b k can be determined by applying the boundary conditions
with respect to y in PDS (7.43)

a k = β k e − λ k b b k = α k − β k e λ k b

−

− α k

,

λ k b .

(7.46)

− 2sh λ k b 2sh

Here

a
0 −

k + 1 q

2
a

q
2 (

2 ( − 1 )

x 2 a 2 α k = − ) cos λ k x d x = ,

3
k

a λ

a
0 −

k + 1

2
a

1
2 (

2 ( − 1 ) ( pb + q )

x 2 a 2 β k = − )( pb + q ) cos λ k x d x = .

3
k

a λ

Finally, we obtain the solution of PDS (7.42) by substituting Eq. (7.46) into
Eq. (7.45) and using the relation between v ( x , y ) and u ( x , y )

p
2 y

q
2

x 2 a 2 u = v ( x , y )+( − ) +

∞
k = 0

k

( − )

1

= 2

λ k b [(

pb + q ) sh λ k ( y − b ) − q sh λ k y ] cos λ k x

3
k sh

λ

1
2 (

x 2 a 2 + py + q )( − ) .