Mathematical Tools for Physics

10—Partial Differential Equations 305

The original potential on the surface was neither even nor odd, but except for the constant average value,
itisan odd function.

z= 0 : V(x,y) =

1

2

V 0 +

{

+V 0 / 2 ( 0 < x < L)

−V 0 / 2 (L < x < 2 L)

V(x+ 2L,y) =V(x,y) (42)

Solve the potential problem for the constantV 0 / 2 and you have a constant. Solve it for the remaining odd
function on the boundary and you should expect an odd function forV(x,z). If you make these observations
beforesolving the problem you can save yourself some algebra, as it will lead you to the form of the solution
faster.
The potential is periodic on thex-yplane, so periodic boundary conditions are the appropriate ones. You
can express these in more than one way, taking as a basis for the expansion either complex exponentials or sines
and cosines.

enπix/L, n= 0,± 1 ,...

or the combination cos(nπx/L), n= 0, 1 ,... sin(nπx/L), n= 1, 2 ,...

(43)

For a random problem with no special symmetry the exponential choice typically leads to easier integrals. In this
case the boundary condition has some symmetry that you can take advantage of: it’s almost odd. The constant
term in Eq. ( 27 ) is then= 0element of the cosine set, and that’s necessarily orthogonal to all the sines. For
the rest, you do the expansion

{

+V 0 / 2 ( 0 < x < L)

−V 0 / 2 (L < x < 2 L)

=

∑∞

1

ansin(nπx/L)

The odd term in the boundary condition ( 42 ) is necessarily a sum of sines, with no cosines. The cosines are
orthogonal to an odd function. See problem 11.

More Electrostatic Examples
Specify the electric potential in thex-y plane to be an array, periodic in both thex and they-directions.
V(x,y,z= 0)isV 0 on the rectangle ( 0 < x < a, 0 < y < b) as well as in the darkened boxes in the picture; it
is zero in the white boxes. What is the potential forz > 0?