Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONS


the corresponding homogeneous equation (i.e. Laplace’s equation) insideVbut


is adjusted in such a way that the sumG(r,r 0 ) equals zero onS. The Green’s


functionG(r,r 0 ) is still a solution of (21.82) since


∇^2 G(r,r 0 )=∇^2 F(r,r 0 )+∇^2 H(r,r 0 )=∇^2 F(r,r 0 )+0=δ(r−r 0 ).

The functionF(r,r 0 ) is called thefundamental solutionand will clearly take

different forms depending on the dimensionality of the problem. Let us first


consider the fundamental solution to (21.82) in three dimensions.


Find the fundamental solution to Poisson’s equation in three dimensions that tends to zero
as|r|→∞.

We wish to solve


∇^2 F(r,r 0 )=δ(r−r 0 ) (21.87)

in three dimensions, subject to the boundary conditionF(r,r 0 )→0as|r|→∞.Sincethe
problem is spherically symmetric aboutr 0 , let us consider a large sphereSof radiusR
centred onr 0 , and integrate (21.87) over the enclosed volumeV. We then obtain


V

∇^2 F(r,r 0 )dV=


V

δ(r−r 0 )dV=1, (21.88)

sinceVencloses the pointr 0. However, using the divergence theorem,


V

∇^2 F(r,r 0 )dV=


S

∇F(r,r 0 )·nˆdS , (21.89)

wherenˆis the unit normal to the large sphereSat any point.
Since the problem is spherically symmetric aboutr 0 , we expect that
F(r,r 0 )=F(|r−r 0 |)=F(r),


i.e. thatFhas the same value everywhere onS. Thus, evaluating the surface integral in
(21.89) and equating it to unity from (21.88), we have§


4 πr^2

dF
dr



∣∣


r=R

=1.


Integrating this expression we obtain


F(r)=−

1


4 πr

+constant,

but, since we requireF(r,r 0 )→0as|r|→∞, the constant must be zero. The fundamental
solution in three dimensions is consequently given by


F(r,r 0 )=−

1


4 π|r−r 0 |

. (21.90)


This is clearly also the full Green’s function for Poisson’s equation subject to the boundary
conditionu(r)→0as|r|→∞.


Using (21.90) we can write down the solution of Poisson’s equation to find,

§A vertical bar to the right of an expression is a common alternative to enclosing the expression in
square brackets; as usual, the subscript shows the value of the variable at which the expression is
to be evaluated.
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