We seek a spherically symmetric solution to the wave equation for
given initial conditions.  Those conditions are:

phi(x,0)=0

phi'(x,0)=f(x)  (where ' means derivative with respect to t).

Theorem:  In 3 dimensions, phi(x,t) is completely determined by the 
values of f on the past light cone of (x,t).

Theorem:  In 2 dimensions, phi(x,t) depends on the values of f both
on *and inside* the past light cone of (x,t).

More precisely:

Let M(x,t) be the mean value of f on the sphere of radius t around x.

Then

Theorem:  In 3 dimensions, phi(x,t) is given by t M(x,t).  

Theorem:  In 2 dimensions, phi(x,t) is given by an integral, where
s ranges from 0 to t and the integrand is:

s M(x,s) / Sqrt[t^2-s^2]  

In other words, in 2 dimensions, the mean value of f on every sphere
of every radius from 0 to t contributes to the solution, while in 3
dimensions, only the sphere of radius t contributes.  In particular,
an initial disturbance at the origin can have effects at x long after
the initial wave crest has passed.

INTUITION:



Given initial data for a 2-dimensional wave, we can create initial
data for a 3-dimensional wave by using the same data and making it
independent of the third coordinate, which I'll call z.

Now if we solve the 3-dimensional problem, we should get a solution
independent of z; restricting to the plane, we've solved our 2-dimensional
problem.

Under this operation, if our initial data are concentrated near the origin
for the 2D-problem, they'll be concentrated all along the z-axis for the
3-D problem.  So from every point along the z-axis, we get an expanding
3-dimensional sphere of non-zero wave. 

Now consider a point P in the plane.  Every one of our vertical array
of expanding spheres will eventually pass through point P.  That's why
(if I've got this right) there will be ongoing non-zero wave values at
point P (and explains exactly why it's everything *inside* the past
light cone, not just *on* the light cone, that matters at a given 
event.