Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PDES: SEPARATION OF VARIABLES AND OTHER METHODS

and, in a similar way, we might wish to choose∂G(r,r 0 )/∂n= 0 in the solution of

Neumann problems. However, in general this isnotpermitted since the Green’s

function must obey the consistency condition
∫

S

∂G(r,r 0 )

∂n

dS=

∫

S

∇G(r,r 0 )·nˆdS=

∫

V

∇^2 G(r,r 0 )dV=1.

The simplest permitted boundary condition is therefore

∂G(r,r 0 )

∂n

=

1

A

forronS,

whereAis the area of the surfaceS; this defines aNeumann Green’s function.

If we require∂u(r)/∂n=f(r)onS, the solution to Poisson’s equation is given by

u(r 0 )=

∫

V

G(r,r 0 )ρ(r)dV(r)+

1

A

∫

S

u(r)dS(r)−

∫

S

G(r,r 0 )f(r)dS(r)

=

∫

V

G(r,r 0 )ρ(r)dV(r)+〈u(r)〉S−

∫

S

G(r,r 0 )f(r)dS(r), (21.106)

where〈u(r)〉Sis the average ofuover the surfaceSand is a freely specifiable con-

stant. For Neumann problems in which the volumeVis bounded by a surfaceSat

infinity, we do not need the〈u(r)〉Sterm. For example, if we wish to solve a Neu-

mann problem outside the unit sphere centred at the origin thenr>ais the region

Vthroughout which we require the solution; this region may be considered as be-

ing bounded by two disconnected surfaces, the surface of the sphere and a surface

at infinity. By requiring thatu(r)→0as|r|→∞,theterm〈u(r)〉Sbecomes zero.

As mentioned above, much of our discussion of Dirichlet problems can be

taken over into the solution of Neumann problems. In particular, we may use the

method of images to find the appropriate Neumann Green’s function.

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

condition∂u/∂n=f(φ)on|r|=a,with

∫ 2 π

0 f(φ)dφ=0as required by the consistency

condition (21.105).

Let us assume, as in Dirichlet problems with this geometry, that a single image charge is
placed outside the circle at

r 1 =

a^2

|r 0 |^2

r 0 ,

wherer 0 is the position of the source inside the circle (see equation (21.100)). Then, from
(21.99), we have the useful geometrical result

|r−r 1 |=

a

|r 0 |

|r−r 0 | for|r|=a. (21.107)

Leaving the strengthqof the image as a parameter, the Green’s function has the form

G(r,r 0 )=

1

2 π

(

ln|r−r 0 |+qln|r−r 1 |+c

)

. (21.108)