Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PDES: SEPARATION OF VARIABLES AND OTHER METHODS

Therefore, in order thatG=0at|r|=a, the strength of the image charge must be
−a/|r 0 |. Consequently, the Dirichlet Green’s function for the exterior of the sphere is

G(r,r 0 )=−

1

4 π|r−r 0 |

+

a/|r 0 |

4 π|r−(a^2 /|r 0 |^2 )r 0 |

.

For a less formal treatment of the same problem see exercise 21.22.

If we seek solutions to Poisson’s equation in theinteriorof a sphere then the

above analysis still holds, butrandr 0 are now inside the sphere and the image

r 1 lies outside it.

For two-dimensional Dirichlet problems outside the circle|r|=a, we are led

by arguments similar to those employed previously to use the same image point

as in the three-dimensional case, namely

r 1 =

a^2

|r 0 |^2

r 0. (21.100)

As illustrated below, however, it is usually necessary to take the image strength

as−1 in two-dimensional problems.

Solve Laplace’s equation in the two-dimensional region|r|≤a, subject to the boundary

conditionu=f(φ)on|r|=a.

In this case we wish to find the Dirichlet Green’s function in the interior of a disc of
radiusa, so the image charge must lie outside the disc. Taking the strength of the image
to be−1, we have

G(r,r 0 )=

1

2 π

ln|r−r 0 |−

1

2 π

ln|r−r 1 |+c,

wherer 1 =(a^2 /|r 0 |^2 )r 0 lies outside the disc, andcis a constant that includes the strength
of the image charge and does not necessarily equal zero.
Since we requireG(r,r 0 )=0when|r|=a, the value of the constantcis determined,
and the Dirichlet Green’s function for this problem is given by

G(r,r 0 )=

1

2 π

(

ln|r−r 0 |−ln

∣∣

∣

∣r−

a^2

|r 0 |^2

r 0

∣∣

∣

∣−ln

|r 0 |

a

)

. (21.101)

Using plane polar coordinates, the solution to the boundary-value problem can be written
as a line integral around the circleρ=a:

u(r 0 )=

∫

C

f(r)

∂G(r,r 0 )

∂n

dl

=

∫ 2 π

0

f(r)

∂G(r,r 0 )

∂ρ

∣

∣

∣

∣

ρ=a

adφ. (21.102)

The normal derivative of the Green’s function (21.101) is given by

∂G(r,r 0 )

∂ρ

=

r

|r|

·∇G(r,r 0 )

=

r

2 π|r|

·

(

r−r 0

|r−r 0 |^2

−

r−r 1

|r−r 1 |^2

)

. (21.103)