Carl wrote:
> What I'm wanting is just a one time kick (at t=0) to occure at x=0, y=0.
> The ripples sould head outward and the amplitude of the oscillation at x=0,
> y=0 should decay back to zero.  It's the solution to the 2D Wave Equation
> with the initial state defined as a Dirac Delta Function.  It's probably
> been nearly 20 years since I saw the solution to this but it may have made
> use of Bessel functions.

I'm a little too lazy to solve it myself right now, but hopefully my 
writing this will remind me to work on it.

I know that in 3-D, for spherical waves, it's given by:

amplitude = (A/abs(r))*exp(i*k.r)

(Looks better on paper). Basically, A is just some constant. r is the 
vector from the starting point. k is the wave number (as a vector in 
3-D), i is the sqrt of -1, and the . between k and r means a dot product.

Now after writing this, I'm not too sure what you want. My one above (in 
3-D), gives the result at a given instant in time, and is only spatially 
dependent. As such, the "decay" is outwards in space, not in time.

It was solved without any delta functions - just a solution to the wave 
equation assuming spherical symmetry, and harmonic solutions.

I think I just had an idea on what you're looking for. If I restrict 
myself to 1-D, and introduce a pulse (as the source - your delta 
function here), that pulse will travel outwards, become of finite 
amplitude, and will spread over time. The pulse actually becomes a 
Gaussian (as the limit of a Gaussian is a delta function).

Looking in a physics text book, I see one solution (assuming the initial 
disturbance was a Gaussian - not delta) as:

amplitude = (a*sqrt(2*pi*(1 + (t/T)^2)))^(-1) *
             exp(-(x - v*t)^2/(2*a^2*(1 + (t/T)^2)))

Looks nicer on paper too. The first "term" is actually in the 
denominator (hence the negative power). The second "term" is all in the 
exponential.

So that's a traveling wave in 1-D. t is time, x is distance from source, 
v is the velocity, a (or a^2) is the spread of the initial Gaussian. T 
is a constant that I can't quite figure out, but is dependent on certain 
parameters involving the wave packet. I'll just do the engineering hand 
wave and suggest you just experiment with various constants for T.

In case the above does not look familiar write it out with t=0, that 
should look like a clear Gaussian. The first "term" is the amplitude, 
the second is the spread. Now note that when t is finite and greater 
than 0, replace a with a*sqrt(1 + (t/T)^2) and consider it to be the 
"new" a. Again, you get a Gaussian, but with a bigger a (hence smaller 
amplitude and wider spread).

(Oh, and BTW, according to my text book, if the initial disturbance were 
a true delta function, (i.e. a = 0), no spreading would occur, so just 
set a to be small).

Phew! I feel bad I had to look up the physics book and not derive it 
from pure math. It's sad how often we rely on the sciences/engineering 
to solve math problems, and not the other way round.

For 2-D, just replace x with sqrt(x*x + y*y) is my guess. Not entirely 
sure, let me know if that works.

Bessel functions, BTW, usually arise in cylindrical geometries.

--
Mueen Nawaz

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