Potential Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 393

0 can be completely neutral-
ized by the electric fields due to the point electric charges of capacity

The electrical potential on the boundary y = 0and x =

ε , ε and − ε

at M 1 ( , respectively, which are the

symmetric points of M 0 . The electric potential of the additional field at point M is

x 0 ,− y 0 , z 0 ) , M 2 ( − x 0 , y 0 , z 0 ) and M 3 ( − x 0 ,− y 0 , z 0 )

1
r M 1 M +

1
4

1
r M 2 M −

1
r M 3 M

g ( M , M 0 )= .

π

1
4

1
r M 0 M

Clearly, Δ g = 0, g | ∂Ω = . Thus the Green function is

π

1
4

1
r M 0 M −

G ( M , M 0 )= g ( M , M 0 )

π

1
r M 0 M −

1
4

1
r M 1 M −

1
r M 2 M +

1
r M 3 M

= .

π

It can be used to obtain the solution of

= ( ) , ∈ Ω , > , > , − ∞ < < + ∞ ,

Δ u F M M x 0 y 0 z

u | ∂Ω = f ( M ) .

Example 4. By using the method of Green functions, find the solution of
Δ

x 2 y 2 R 2

u = 0 , + < ,

(7.117)

u | x 2 = F ( x , y ) .

y 2 R 2 + =

Solution. In a two-dimensional plane domain, from Section 7.6 the Green function
is

1
2

1
r M 0 M −

( ,

G M M 0 )= ln g ( M , M 0 ) ,

π

where r M 0 M is the distance between M 0 and M and the g ( M , M 0 ) satisfies

⎧
⎨

Δ g = 0 , M and M 0 areinsidethecircle r = R ,

1
2

1
r M 0 M .

(7.118)

⎩ g | r = R = ln

π

To obtain the solution of PDS (7.118), consider the symmetric point

( , ) ( , ) =

M 1 r 1 cos θ r 1 sin θ of M 0 r 0 cos θ r 0 sin θ with respect to the circle r R 0 0 0 0

R 2 . Note that

=

(Fig. 7.3) such that r 0 r 1

r 0 + r 1 +

r M 0 M = r 2 − 2 rr 0 cos ( θ − θ 0 ) , r M 1 M = r 2 − 2 rr 1 cos ( θ − θ 0 ) .