10—Partial Differential Equations 255does one EM wave get through while the other one doesn’t? I won’t solve the whole electromagnetic
radiation problem here, but I’ll look at the static analog to get some general idea of what’s happening.
−L 0 L 2 L x
z
y V^0
Arrange a set of conducting strips in thex-yplane and with insulation between them so that
they don’t quite touch each other. Now apply voltageV 0 on every other one so that the potentials are
alternately zero andV 0. This sets the potential in thez= 0plane to be independent ofyand
z= 0 : V(x,y) =
{
V 0 ( 0 < x < L)
0 (L < x < 2 L) V(x+ 2L,y) =V(x,y), allx,y (10.35)
What is then the potential above the plane,z > 0? Above the planeV satisfies Laplace’s equation,
∇^2 V=
∂^2 V
∂x^2
+
∂^2 V
∂y^2
+
∂^2 V
∂z^2
= 0 (10.36)
The potential is independent ofyin the plane, so it will be independent ofyeverywhere. Separate
variables in the remaining coordinates.
V(x,z) =f(x)g(z) =⇒
d^2 f
dx^2
g+f
d^2 g
dz^2
= 0 =⇒
1
f
d^2 f
dx^2
+
1
g
d^2 g
dz^2
= 0
This is separated as a function ofxplus a function ofy, so the terms are constants.
1
f
d^2 f
dx^2
=−α^2 ,
1
g
d^2 g
dz^2
= +α^2 (10.37)
I’ve chosen the separation constant in this form because the boundary condition is periodic inx, and
that implies that I’ll want oscillating functions there, not exponentials.
f(x) =eiαx and f(x+ 2L) =f(x)
=⇒e^2 Liα= 1, or 2 Lα= 2nπ, n= 0,± 1 ,± 2 ,...
The separated solutions are then
f(x)g(z) =enπix/L
(
Aenπz/L+Be−nπz/L
)
(10.38)
The solution forz > 0 is therefore the sum
V(x,z) =
∑∞
n=−∞