Equation for a raindrop falling on a puddle for a mechanical art project Given a raindrop falling on a puddle, is there an equation that describes the resulting motion?
I found this article that describes it well, so I was thinking it could be
$$z = \frac{A\sin(kx+wt)}{x}$$ rotated around the z-axis. However I'm not sure how I could verify this.
Could the wave equation also work for this?
$$ \frac{\partial^{2}z}{\partial t^2} = c^2 \left(\frac{\partial^{2}z}{\partial x^2}+\frac{\partial^{2}z}{\partial y^2} \right )$$
The background for this question is I'm working on simulating water waves using motors
connected to tiles for an art project. I want to understand how the continuous displacement of water translates into discrete instructions for servo motors. Thank you.

 A: What you are doing is a very simple model, where a drop is considered a point-like (1) and instantaneous (2) perturbation, and the resulting waves are treated as two-dimensional surface waves (3) (although there may be also waves going into the depth).
Provided that we agree to use all these approximations, you still cannot simply rotate a plane wave solution, but rather need to use the wave equations in polar coordinates. More specifically, the right-hand side of your equation becomes:
$$
\nabla^2 z(x,y)=\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2}=
\frac{\partial^2 z}{\partial r^2} + \frac{1}{r}\frac{\partial^2 z}{\partial r}+
\frac{1}{r^2}\frac{\partial^2 z}{\partial \theta^2},
$$
and the solution will be in terms of Bessel functions (the books by Abramovitz&Stegun and Gradshtein&Ryzhik are comprehensive references on Bessel functions, although the basics can be found in the books on electricity&Magnetism or Quantum mechanics).
A: The wave equation in polar coordinates is (in PDE shorthand):
$$z_t=c^2\left(z_{rr}+\frac{2z_r}{r}\right)\tag{1}$$
So we're looking for a function $z(r,t)$, which we assume ('Ansatz') to be of the form:
$$z(r,t)=R(r)T(t)$$
Inserting into $(1)$ and reworking gives:
$$\frac{1}{c^2}RT'=TR''+\frac{2}{r}TR'$$
$$\frac{T'}{c^2T}=\frac{R''}{R}+\frac{2R'}{rR}=-k^2$$
where $-k^2$ is a separation constant (a Real Number). This gives us two ODEs:
$$\frac{T'}{c^2T}=-k^2\tag{2}$$
$$\frac{R''}{R}+\frac{2R'}{rR}=-k^2\tag{3}$$
$(3)$ reworked gives
$$rR''+2R'+k^2rR=0$$
Now substitute and insert:
$$\rho(r)=rR(r)$$
$$\rho'=R+rR'$$
$$\rho''=2R'+rR''$$
$$\rho''(r)+k^2\rho(r)=0$$
To get:
$$\rho(r)=c_1\sin kr+c_2\cos kr$$
And:
$$R(r)=\frac{c_1\sin kr}{r}+\frac{c_2\cos kr}{r}$$
Now note that:
$$\lim_{r \to 0}\left[\frac{c_2\cos kr}{r}\right] \to +\infty \Rightarrow c_2=0$$
So that:
$$R(r)=\frac{c_1\sin kr}{r}$$
Solving $(2)$ is easy:
$$\frac{T'}{c^2T}=-k^2$$
$$T(t)=c_3\exp\left(-c^2k^2 t\right)$$
So that we can write:
$$z(r,t)=B\exp\left(-c^2k^2 t\right)\frac{\sin kr}{r}$$

Now, in order to determine both $B$ and $k$ we need an initial condition (IC, $t=0$) and a boundary condition (BC).
For IC we could use a sharp, Gaussian-style $z(r,0)=f(0)$ function.
But finding a suitable BC is harder. From the OP's dispersion graphs, we can see that $z(0,t)=0$ which means:
$$R(0)=0$$
Unfortunately that doesn't work because:
$$\lim_{r\to 0}\left[\frac{\sin kr}{r}\right]=k$$
and not zero.
