Mathematical Methods for Physics and Engineering : A Comprehensive Guide

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONS

ˆn nˆ

nˆ

V

V S

S 1

S 2

(a) (b)

Figure 21.11 Surfaces used for solving Poisson’s equation in different

regionsV.

where on the RHS it is common to write, for example,∇ψ·nˆdSas (∂ψ/∂n)dS.

The expression∂ψ/∂nstands for∇ψ·nˆ, the rate of change ofψin the direction

of the unit outward normalnˆto the surfaceS.

The Green’s function for Poisson’s equation (21.80) must satisfy

∇^2 G(r,r 0 )=δ(r−r 0 ), (21.82)

wherer 0 lies inV. (As mentioned above, we may think ofG(r,r 0 )asthesolution

to Poisson’s equation for a unit-strength point source located atr=r 0 .) Let us

for the moment impose no boundary conditions onG(r,r 0 ).

If we now letφ=u(r)andψ=G(r,r 0 ) in Green’s theorem (21.81) then we

obtain
∫

V

[

u(r)∇^2 G(r,r 0 )−G(r,r 0 )∇^2 u(r)

]

dV(r)

=

∫

S

[

u(r)

∂G(r,r 0 )

∂n

−G(r,r 0 )

∂u(r)

∂n

]

dS(r),

where we have made explicit that the volume and surface integrals are with

respect tor. Using (21.80) and (21.82) the LHS can be simplified to give
∫

V

[u(r)δ(r−r 0 )−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). (21.83)

Sincer 0 lies within the volumeV,
∫

V

u(r)δ(r−r 0 )dV(r)=u(r 0 ),

and thus on rearranging (21.83) the solution to Poisson’s equation (21.80) can be