Potential Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 340

It can be transformed into an equation with constant coefficients by a variable trans-
formation of r

e t . Its general solution can thus be readily obtained as

=

c 0 +

d 0 ln r , n = 0 ,

( )=

R n r

c n r n d n r − n

+ , n = 1 , 2 , ··· ,

where the c n and the d n are constants. Applying | R ( 0 ) | < ∞ yields d n = 0, so that

c n r n R n ( r )= , n = 0 , 1 , 2 , ··· .

Thus

∞
n = 1 r n

a 0
2 +

u ( r , θ )= ( a n cos n θ + b n sin n θ ) . (7.15)

Applying the boundary condition u | r = a = f ( θ ) yields

⎧
⎨

1
a n

2π
0

a n = f ( θ ) cos n θ d θ ,

π

(7.16)

⎩

1
a n

2π
0

b n = f ( θ ) sin n θ d θ , n = 0 , 1 , 2 , ··· .

π

Remark 1. Note that

n = 1 z n cos n ( θ − θ )= − 1 + 2 ∞

∞

n = 0 z n cos n ( θ − θ )

1 + 2

2Re ∞

n = 0 z n e i n ( θ − θ )

2Re

1

= − 1 + = − 1 +

z e i ( θ − θ ) 1 −

2Re 1

( θ − θ )+ ( θ − θ ) − z cos i z sin

= − 1 +

( θ − θ )+

1 − 2 z cos z 2

( θ − θ )

2 − 2 z cos

= − 1 +

( θ − θ )+ z 2 1 − 2 z cos

z 2

1 −

=

( θ − θ ) ,

| z | < 1 .

1 + z 2 − 2 z cos

The u ( r , θ ) in Eq. (7.15) becomes

2π

a 2 r 2

1
2

−

( θ ) θ . u ( r , θ )= f d (7.17)

a 2 r 2 ( θ − θ ) π + − 2 ar cos 0

This is called the Poisson formula of internal problems in a circle .

Remark 2 . An external problem in a circle reads
⎧
⎨

2 u 2 u

∂θ

∂ 1

r ∂

u

r 2 ∂

1

r 2 + r +

2 =

, < < + ∞ ,

0 a r

∂ ∂

⎩

u | r = a = f ( θ ) , f ( θ + 2 π )= f ( θ ) .