Mathematical Methods for Physics and Engineering : A Comprehensive Guide

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONS

(ii) Since all the image sources lie outsideV, the fundamental solution cor-

responding to each source satisfies Laplace’s equationinsideV. Thus we

may add the fundamental solutionsF(r,rn) corresponding to each image

source to that corresponding to the single source insideV, obtaining the

Green’s function

G(r,r 0 )=F(r,r 0 )+

∑N

n=1

qnF(r,rn).

(iii) Now adjust the positionsrnand strengthsqnof the image sources so

that the required boundary conditions are satisfied onS. For a Dirichlet

Green’s function we requireG(r,r 0 )=0forronS.

(iv) The solution to Poisson’s equation subject to the Dirichlet boundary

conditionu(r)=f(r)onSis then given by (21.86).

In general it is very difficult to find the correct positions and strengths for the

images, i.e. to make them such that the boundary conditions onSare satisfied.

Nevertheless, it is possible to do so for certain problems that have simple geometry.

In particular, for problems in which the boundarySconsists of straight lines (in

two dimensions) or planes (in three dimensions), positions of the image points

can be deduced simply by imagining the boundary lines or planes to be mirrors

in which the single source inV(atr 0 ) is reflected.

Solve Laplace’s equation∇^2 u=0in three dimensions in the half-spacez> 0 , given that

u(r)=f(r)on the planez=0.

The surfaceSboundingVconsists of thexy-plane and the surface at infinity. Therefore,
the Dirichlet Green’s function for this problem must satisfyG(r,r 0 )=0onz=0and
G(r,r 0 )→0as|r|→∞. Thus it is clear in this case that we require one image source at a
positionr 1 that is the reflection ofr 0 in the planez= 0, as shown in figure 21.12 (so that
r 1 lies inz<0, outside the region in which we wish to obtain a solution). It is also clear
that the strength of this image should be−1.
Therefore by adding the fundamental solutions corresponding to the original source
and its image we obtain the Green’s function

G(r,r 0 )=−

1

4 π|r−r 0 |

+

1

4 π|r−r 1 |

, (21.94)

wherer 1 is the reflection ofr 0 in the planez=0,i.e.ifr 0 =(x 0 ,y 0 ,z 0 )thenr 1 =(x 0 ,y 0 ,−z 0 ).
ClearlyG(r,r 0 )→0as|r|→∞as required. AlsoG(r,r 0 )=0onz= 0, and so (21.94) is
the desired Dirichlet Green’s function.
The solution to Laplace’s equation is then given by (21.86) withρ(r)=0,

u(r 0 )=

∫

S

f(r)

∂G(r,r 0 )

∂n

dS(r). (21.95)

Clearly the surface at infinity makes no contribution to this integral. The outward-pointing
unit vector normal to thexy-plane is simplyˆn=−k(wherekis the unit vector in the
z-direction), and so

∂G(r,r 0 )

∂n

=−

∂G(r,r 0 )

∂z

=−k·∇G(r,r 0 ).