We can use the same method to construct the interior Green functions

for the sphere and circle.

Kirchho¬ vs. Huygens

6.4.4

Even if we do not have a Green function tailored for the speci¬c region in

which were are interested, we can still use the whole-space Green function

to convert the di¬erential equation into an integral equation, and so make

progress. An example of this technique is provided by Kirchho¬™s partial

justi¬cation of Huygens™ construction.

The Green function G(r, r ) for the elliptic Helmholtz equation

2

+ κ2 )G(r, r ) = δ 3 (r ’ r )

(’ (6.167)

in R3 is given by

d3 k eik·(r’r ) 1

e’κ|r’r | .

= (6.168)

(2π)3 k 2 + κ2 4π|r ’ r |

Exercise: Perform the k integration and con¬rm this.

For solutions of the wave equation with e’iωt time dependence, we want

a Green function such that

ω2

2

G(r, r ) = δ 3 (r ’ r ),

’ ’ (6.169)

c2

and so we have to take κ2 negative. We therefore have two possible Green

functions

1

e±ik|r’r | ,

G± (r, r ) = (6.170)

4π|r ’ r |

where k = |ω|/c. These correspond to taking the real part of κ2 negative, but

giving it an in¬nitesimal imaginary part, as we did when discussing resolvent

operators in chapter 5. If we want outgoing waves, we must take G ≡ G+ .

180 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS

Now suppose we want to solve

2

’ k2 )ψ = 0

(’ (6.171)

in an arbitrary region „¦. As before, we use Green™s theorem to write

2

+ k 2 )ψ(r) ’ ψ(r)( 2

+ k 2 )G(r, r ) dn x

G(r, r )( r r

„¦

{G(r, r ) ’ ψ(r) )} · dSr

= r ψ(r) r G(r, r (6.172)

‚„¦

where dSr = n dSr , with n the outward normal to ‚„¦ at the point r. The

left hand side is

ψ(r)δ n (r ’ r ) dn x = ψ(r ), r ∈„¦ (6.173)

„¦

and so

{G(r, r )(n · ’ ψ(r)(n · r ∈ „¦.

ψ(r ) = x )ψ(r) r )G(r, r )} dSr ,

‚„¦

(6.174)

This must not be thought of as solution to the wave equation in terms of an

integral over the boundary, analogous to the solution of the Dirichlet problem

we found earlier. Here, unlike that earlier case, G(r, r ) knows nothing of the

boundary ‚„¦, and so both terms in the surface integral contribute to ψ. We

therefore have a formula for ψ(r) in the interior in terms of both Dirichlet

and Neumann data on the boundary ‚„¦, and giving both over-prescribes the

problem. If we take arbitrary values for ψ and (n · )ψ on the boundary, and

use our formula to compute ψ(r) as r approaches the boundary, then there

is no reason why the resulting ψ(r) should reproduce the assumed boundary

values of ψ and (n · )ψ. If we demand that it does reproduce the boundary

data, then this is equivalent to demanding that the boundary data come from

a solution of the di¬erential equation in a region encompassing „¦.

The mathematical inconsistency of assuming arbitrary boundary data

notwithstanding, this is exactly what we do when we follow Kirchho¬ and

use this formula to provide a justi¬cation of Huygens™ construction as used in

optics. Consider the problem of a plane wave, ψ = eikx , incident on a screen

from the left and passing though the aperture labelled AB in the following

¬gure.

6.4. LAPLACE™S EQUATION 181

„¦

A r™

R

θ

n r

B

Huygens™ construction.

We take the region „¦ to be everything to the right of the obstacle.

The Kirchho¬ approximation consists of assuming that the values of ψ and

(n · )ψ on the surface AB are eikx and ’ikeikx , the same as they would be

if the obstacle were not there, and that they are identically zero on all other

parts of the boundary. In other words, we completely ignore any scattering

by the material in which the aperture resides. We can then use our formula

to estimate ψ in the region to the right of the aperture. If we further set

(r ’ r ) ik|r’r |

r G(r, r ) ≈ ik e , (6.175)

|r ’ r |2

which is a good approximation provided we are more than a few wavelengths

away from the aperture, we ¬nd

eik|r’r |

k

ψ(r ) ≈ (1 + cos θ)dSr . (6.176)

|r ’ r |

4πi aperture

Thus, each part of the wavefront on the surface AB acts as a source for the

di¬racted wave in „¦.

This result, although still an approximation, provides two substantial

improvements to the na¨ form of Huygens™ construction as presented in

±ve

elementary courses:

182 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS

i) There is factor of (1 + cos θ) which suppresses backward propagating

waves. The traditional exposition of Huygens construction takes no

notice of which way the wave is going, and so provides no explanation

as to why a wavefront does not act a source for a backward wave.

ii) There is a factor of i’1 = e’iπ/2 which corrects a 90—¦ error in the phase

made by the na¨ Huygens construction. For two-dimensional slit

±ve

geometry we must use the more complicated two-dimensional Green

function (it is a Bessel function), and this provides an e’iπ/4 factor

which corrects for the 45—¦ phase error that is manifest in the Cornu