Potential Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 341

It can be solved by following the same approach in Example 1.
Here d 0 = 0and c n = 0in

c 0

+ d 0 ln r , n = 0 ,

R n ( r )=

c n r n d n r − n

+ , n = 1 , 2 , ··· .

Thus the r n in Eq. (7.15) should be replaced by r − n . Finally, the solution is

2π

a 2 r 2 1

2

−

( θ ) θ . u ( r , θ )= − f d

a 2

+

r 2

− ( θ − θ )

π 2 ar cos

0

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

Remark 3 . The general form of homogeneous Euler equations of second order is

x 2 y + pxy +

qy = 0 , (7.18)

where p and q are constants. It reduces to an equation with constant coefficients by
a variable transformation of x e t

=

D ( D − 1 ) y + pDy + qy = 0 , (7.19)

d
d t . Its characteristic equation can be obtained by replacing D and y in
Eq. (7.19) by the characteristic roots r and 1, respectively.

=

where D

r ( r − 1 )+ pr + q = 0 .

Remark 4 . The result in Example 1 can also be used to solve problems in a semi-
circular domain by using the method of continuation. Consider two-dimensional
steady heat conduction, for example, in a semi-circular plate of radius a .Thetem-
perature is u with T 0 as a constant on the semi-circle of the

boundary and zero on the other part of the boundary. Thus, by an odd continuation
of boundary value, the temperature u

( a , θ )= T 0 θ ( π − θ )

( r , θ ) satisfies

Δ

u ( r , θ )= 0 , 0 < r < a , − π < θ < π ,

u ( a , θ )= f ( θ ) ,

T 0 θ ( π − θ ) ,

< θ < π ,
T 0 θ ( π + θ ) , − π < θ <

0

where f ( θ )= Note that f ( θ ) is an odd function of

0 .

θ so that a n = 0 in Eq. (7.16). Thus

∞
n = 1 b n r n sin n θ .

u ( r , θ )=

Applying the boundary condition at r = a yields

∞
n = 1 b n a n sin n θ = T 0 θ ( π − θ ) ,