Spherical Waves-Strength at close distances If the amplitude dies off as the radius squared, what happens in areas very close to the source? It would have nearly infinite strength. How is this treated?
 A: The closest analogy that is readily talked about in spherical wave terms is an antenna. 
To analyse the problem fully, you would need to solve Maxwell's equations in several, distict regions of space. The spherical wave is a often a good model in the homogeneous space outside the antenna's hardware. Often this is good enough to tell you what you want - e.g. received power or electric / magnetic fields in terms of the antenna's output power. The spherical waves are theoretically output by infinitely small / thin currents and charge distributions.
However, if you want to know in detail what is going on in the antenna's hardware, you'll have to solve Maxwell's equations inside the conducting hardware and match this separate solution through standard boundary conditions to the superposition of spherical waves in the free space outside the antenna. Inside the hardware, you have moving currents spread over nonzero volumes, so you don't get the divergences that you otherwise would if you assumed the spherical wave solution applied everywhere. 
In other words - you leave off the spherical wave model before it diverges and shift to a fuller model of the antenna's hardware.
The simplest way to see all this in action is an electrostatics problem: calculate the electric field of a spherical charged ball wherein the charge is uniformly spread. This is a spherically symmetric problem, so clearly the electric field is everywhere radial. We can most readily apply Gauss's law for electricity over spherical surfaces concentric with the charged ball. Suppose that this spread charge is $Q$.
So let's begin with a surface of radius $r$ where $r > R$, where $R$ is the charged ball's radius. The electric field, by symmetry, is normal to the surface everywhere, and of constant magnitude $E(r)$ (where $E(r)$ is to be calculated) on the surface. The surface's area is $4\pi r^2$ So the flux of the electric field through the sphere is:
$\oint \mathbf{E} . \hat{\mathbf{n}}\, dS = 4\pi r^2 E(r)$
Gauss's law says:
$\oint \mathbf{E} . \hat{\mathbf{n}}\, dS = \frac{q}{\epsilon_0}$
where $q(r)$ is the charge inside the sphere. But, because $r > R$, $q(r) = Q$ is constant. So, outside the charged sphere we get:
$4\pi r^2 E(r) = \frac{Q}{\epsilon_0}$ i.e. $E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
which is the "divergent" behavior you're worried about. Moreover, note that this is exactly the same expression holding for a point charge (outside the sphere, we can't tell how big the sphere is from electrostatic measurements alone).
However, in the second case, let's put our sphere of radius $r$ inside the charged ball. Now, the charge inside the sphere depends on $r$: if $Q$ is spread evenly over volume, the fraction of charge enclosed by the sphere scales like $r^3$ (since this is how volume scales). So, if we do the above Gauss's law calculation again we get:
$\oint \mathbf{E} . \hat{\mathbf{n}}\, dS = 4\pi r^2 E(r) = \frac{r^3}{R^3} Q$
i.e.
$E(r) = \frac{Q}{4 \pi \epsilon_0 R^2} r$
So the divergence has vanished!! ($R$ is a constant, the charged sphere's radius). So the field as a function of all $r$ looks like the Mathematica-drawn plot below. The principles for the dynamic situation of antennas are exactly the same: the currents driving the fields are not true threads but are spread out.

A: 
I was just thinking some spherical light source. Like, a light bulb? –  Anthony Jul 22 '13 at 3:32 

For a light bulb there cannot be infinite strength because the power is given by I*V and the bulb has resistance, so the current is limited. The resistance that is heated to give off light will not be giving off light at r=0, for whatever topology you might think.
In any case, the 1/r^2 behavior is valid at distances larger than the size of the source giving off light, and is modified near the boundaries where the light is produced.
