Mathematical Methods for Physics and Engineering : A Comprehensive Guide

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONS

example, in solving Poisson’s equation in two dimensions in the half-spacex> 0

we again require just one image charge, of strengthq 1 =−1, at a positionr 1 that

is the reflection ofr 0 in the linex= 0. Since we requireG(r,r 0 )=0whenrlies

onx= 0, the constant in (21.93) must equal zero, and so the Dirichlet Green’s

function is

G(r,r 0 )=

1

2 π

(

ln|r−r 0 |−ln|r−r 1 |

)

.

ClearlyG(r,r 0 ) tends to zero as|r|→∞. If, however, we wish to solve the two-

dimensional Poisson equation in the quarter spacex>0,y>0, then more image

points are required.

A line charge in thez-direction of charge densityλis placed at some positionr 0 in the

quarter-spacex> 0 ,y> 0. Calculate the force per unit length on the line charge due to

the presence of thin earthed plates alongx=0andy=0.

Here we wish to solve Poisson’s equation,

∇^2 u=−

λ

 0

δ(r−r 0 ),

in the quarter spacex>0,y>0. It is clear that we require three image line charges
with positions and strengths as shown in figure 21.13 (all of which lie outside the region
in which we seek a solution). The boundary condition that the electrostatic potentialuis
zero onx=0andy= 0 (shown as the ‘curve’Cin figure 21.13) is then automatically
satisfied, and so this system of image charges is directly equivalent to the original situation
of a single line charge in the presence of the earthed plates alongx=0andy= 0. Thus
the electrostatic potential is simply equal to the Dirichlet Green’s function

u(r)=G(r,r 0 )=−

λ

2 π 0

(

ln|r−r 0 |−ln|r−r 1 |+ln|r−r 2 |−ln|r−r 3 |

)

,

which equals zero onCand on the ‘surface’ at infinity.
The force on the line charge atr 0 , therefore, is simply that due to the three line charges
atr 1 ,r 2 andr 3. The elecrostatic potential due to a line charge atri,i= 1, 2 or 3, is given
by the fundamental solution

ui(r)=∓

λ

2 π 0

ln|r−ri|+c,

the upper or lower sign being taken according to whether the line charge is positive or
negative, respectively. Therefore the force per unit length on the line charge atr 0 , due to
the one atri,isgivenby

−λ∇ui(r)

∣∣

∣

∣

r=r 0

=±

λ^2

2 π 0

r 0 −ri

|r 0 −ri|^2

.

Adding the contributions from the three image charges shown in figure 21.13, the total
force experienced by the line charge atr 0 is given by

F=

λ^2

2 π 0

(

−

r 0 −r 1

|r 0 −r 1 |^2

+

r 0 −r 2

|r 0 −r 2 |^2

−

r 0 −r 3

|r 0 −r 3 |^2

)

,

where, from the figure,r 0 −r 1 =2y 0 j,r 0 −r 2 =2x 0 i+2y 0 jandr 0 −r 3 =2x 0 i. Thus, in