Mathematical Methods for Physics and Engineering : A Comprehensive Guide

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONS

Using the fact thatr 1 =(a^2 /|r 0 |^2 )r 0 and the geometrical result (21.99), we find that

∂G(r,r 0 )

∂ρ

∣

∣

∣∣

ρ=a

=

a^2 −|r 0 |^2

2 πa|r−r 0 |^2

.

In plane polar coordinates,r=ρcosφi+ρsinφjandr 0 =ρ 0 cosφ 0 i+ρ 0 sinφ 0 j,
and so
∂G(r,r 0 )
∂ρ

∣

∣

∣

∣

ρ=a

=

(

1

2 πa

)

a^2 −ρ^20

a^2 +ρ^20 − 2 aρ 0 cos(φ−φ 0 )

.

On substituting into (21.102), we obtain

u(ρ 0 ,φ 0 )=

1

2 π

∫ 2 π

0

(a^2 −ρ^20 )f(φ)dφ

a^2 +ρ^20 − 2 aρ 0 cos(φ−φ 0 )

, (21.104)

which is the solution to the problem.

21.5.4 Neumann problems

In a Neumann problem we require the normal derivative of the solution of

Poisson’s equation to take on specific values on some surfaceSthat boundsV,

i.e. we require∂u(r)/∂n=f(r)onS,wherefis a given function. As we shall see,

much of our discussion of Dirichlet problems can be immediately taken over into

the solution of Neumann problems.

As we proved in section 20.7 of the previous chapter, specifying Neumann

boundary conditions determines the relevant solution of Poisson’s equation to

within an (unimportant) additive constant. Unlike Dirichlet conditions, Neumann

conditions impose a self-consistency requirement. In order for a solutionuto exist,

it is necessary that the following consistency condition holds:
∫

S

fdS=

∫

S

∇u·nˆdS=

∫

V

∇^2 udV=

∫

V

ρdV, (21.105)

where we have used the divergence theorem to convert the surface integral into

a volume integral. As a physical example, the integral of the normal component

of an electric field over a surface bounding a given volume cannot be chosen

arbitrarily when the charge inside the volume has already been specified (Gauss’s

theorem).

Let us again consider (21.84), which is central to our discussion of Green’s

functions in inhomogeneous problems. It reads

u(r 0 )=

∫

V

G(r,r 0 )ρ(r)dV(r)+

∫

S

[

u(r)

∂G(r,r 0 )

∂n

−G(r,r 0 )

∂u(r)

∂n

]

dS(r).

As always, the Green’s function must obey

∇^2 G(r,r 0 )=δ(r−r 0 ),

wherer 0 lies inV. In the solution of Dirichlet problems in the previous subsection,

we chose the Green’s function to obey the boundary conditionG(r,r 0 )=0onS