Mathematical Methods for Physics and Engineering : A Comprehensive Guide

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONS

y

z

a

−a

B

A

V

x

−a

|r 0 | r 1

+1 r 0

Figure 21.14 The arrangement of images for solving Poisson’s equation

outside a sphere of radiusacentred at the origin. For a charge +1 atr 0 ,the

image pointr 1 is given by (a/|r 0 |)^2 r 0 and the strength of the image charge is

−a/|r 0 |.

By symmetry we expect the image pointr 1 to lie on the same radial line as the original
source,r 0 , as shown in figure 21.14, and sor 1 =kr 0 wherek<1. However, for a Dirichlet
Green’s function we requireG(r−r 0 )=0on|r|=a, and the form of the Green’s function
suggests that we need

|r−r 0 |∝|r−r 1 | for all|r|=a. (21.97)

Referring to figure 21.14, if this relationship is to hold over the whole surface of the
sphere, then it must certainly hold for the pointsAandB. We thus require

|r 0 |−a

a−|r 1 |

=

|r 0 |+a

a+|r 1 |

,

which reduces to|r 1 |=a^2 /|r 0 |. Therefore the image point must be located at the position

r 1 =

a^2

|r 0 |^2

r 0.

It may now be checked that, for this location of the image point, (21.97) is satisfied over
the whole sphere. Using the geometrical result

|r−r 1 |^2 =|r|^2 −

2 a^2

|r 0 |^2

r·r 0 +

a^4

|r 0 |^2

=

a^2

|r 0 |^2

(

|r 0 |^2 − 2 r·r 0 +a^2

)

for|r|=a, (21.98)

we see that, on the surface of the sphere,

|r−r 1 |=

a

|r 0 |

|r−r 0 | for|r|=a. (21.99)