The pole 1/(n ’ 2) is divergent, but independent of position. We can absorb

it, and the ’ ln π, into an undetermined additive constant. Once we have

done this, the limit n ’ 2 can be taken and we ¬nd

1

g(r, r ) = ’ ln |r ’ r | + const., n = 2. (6.154)

2π

We now look at the general interior Dirichlet problem in a region „¦.

n

r

r™

„¦

Interior Dirichlet problem.

We wish to solve ’ 2 • = q(r) for r ∈ „¦ and with •(r) = f (r) for r ∈ ‚„¦.

Suppose we have found a Green function that obeys

2

) = δ n (r ’ r ),

’ r, r ∈ „¦, r ∈ ‚„¦.

r g(r, r g(r, r ) = 0, (6.155)

We can show that g(r, r ) = g(r , r) by the same methods we used for one-

dimensional self-adjoint operators. Next we follow the same strategy that

we used for the heat equation. We use Lagrange™s identity (in this context

called Green™s theorem) to write

dn r g(r, r ) 2 2

’ •(r)

r •(r) r g(r, r )

„¦

dS · {g(r, r ) ’ •(r)

= r •(r) r g(r, r )} ,(6.156)

‚„¦

where dS = n dS, with n the outward normal to ‚„¦. The left hand side is

dn r {’g(r, r )q(r) + •(r)δ n (r ’ r )} ,

L.H.S. =

„¦

dn r g(r, r ) q(r) + •(r ),

=’

„¦

dn r g(r , r) q(r) + •(r ).

=’ (6.157)

„¦

6.4. LAPLACE™S EQUATION 177

On the right hand side, the boundary condition on g(r, r ) makes the ¬rst

term zero, so

R.H.S = ’ dSf (r)(n · r )g(r, r ). (6.158)

‚„¦

Therefore,

g(r , r) q(r) dnr ’ f (r)(n ·

•(r ) = r )g(r, r ) dS. (6.159)

„¦ ‚„¦

In the language of chapter 3, the ¬rst term is a particular integral and the

second (the boundary integral term) is the complementary function.

Exercise: Show that the limit of •(r ) as r approaches the boundary is indeed

f (r ). (Hint: When r, r are very close to it, assume that the boundary can

be approximated by a straight line segment, and so g(r, r ) can be found by

the method of images.)

A similar method works for the exterior Dirichlet problem.

r

„¦

Exterior Dirichlet problem.

Here we seek a Green function obeying

2

) = δ n (r ’ r ), r, r ∈ Rn \ „¦

’ r ∈ ‚„¦. (6.160)

r g(r, r g(r, r ) = 0,

(The notation Rn \ „¦ means the region outside „¦.) We also impose a further

boundary condition by requiring g(r, r ), and hence •(r), to tend to zero as

|r| ’ ∞. The ¬nal formula for •(r) is the same except for the region of

integration and the sign of the boundary term.

The hard part of both the interior and exterior problems is to ¬nd the

Green function for the given domain.

6.4.3 Method of Images

When „¦ is a sphere or a circle we can ¬nd the Green functions by using the

method of images.

178 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS

Consider a circle of radius R.

X

O

B

A

Points inverse with respect to a circle.

Given B outside the circle, and a point X on the circle, we construct A inside,

so that OBX = OXA. We observe that XOA is similar to BOX, and

so

OA OX

= . (6.161)

OX OB

Thus, OA — OB = (OX)2 ≡ R2 , and the points A and B are mutually

inverse with respect to the circle. In particular, the point A does not depend

on which point X was chosen.

Now let AX= ri , BX= r0 and OB= B. Then, using similarity again, we

have

AX BX

= , (6.162)

OX OB

or

R B

=, (6.163)

ri r0

and so

1 R 1

’ = 0. (6.164)

ri B r0

Interpreting the ¬gure as a slice through the centre of a sphere of radius R,

we see that if we put a unit charge at B, then the insertion of an image charge

of magnitude q = ’R/B at A serves to the keep the entire surface of the

sphere at zero potential.

Thus, in three dimensions, and with „¦ the region exterior to the sphere,

we have

1 1 R 1

’

g„¦ (r, rB ) = . (6.165)

4π |r ’ rB | |rB | |r ’ rA |

6.4. LAPLACE™S EQUATION 179

In two dimensions, we ¬nd similarly that

1

g„¦ (r, rB ) = ’ ln |r ’ rB | ’ ln |r ’ rA | ’ ln (|rB |/R) , (6.166)

2π

has g„¦ (r, rB ) = 0 for r on the circle. Thus, this is the Dirichlet Green function