I've been working on trying to find or parametrise the equation of the curve one gets if one drops a mass, say a ball, on a rubber sheet. The shape would look like those popular analogies of gravity and general relativity which use rubber sheets to demonstrate the action of gravity.
To simplify matters, I considered the rubber to be two-dimensional and the weight to be infinitesimally small. In addition, due to my limited knowledge of physics, the forces involved, and differential equations, my assumptions are mostly based on intuition and that limited knowledge of mechanics.
Now, by Hooke's law, the amount that the sheet will stretch is proportional to its extension, so if we consider a point $P$, with coordinates $(x, E(x))$ on our sheet, and consider only the force acting straight down, parallel to the force of gravity, we know that the the amount point $P$ sinks by, from its position without the weight, $E(x)$, will be proportional to that force.
$$F_P = k E_P$$ where E is the vertical "extension", how much point $P$ sinks by, and k is the constant of proportionality.
Moreover, the constant of proportionality will be constant throughout the sheet.
Now, we need to find what that force acting on $P$ is.
We know that the tension in the rubber should be the same everywhere, so at point $P$, the tension will act in the direction parallel to the gradient of the rubber sheet at point $P$. Therefore, to find the downwards force, we must find the vertical component of that tension vector.
This is just a force diagram problem, so solving it gives:
$$F_P = T \sin(\theta)$$ where T is the tension and $\theta$ is equal to $\arctan(E')$, where E' is the derivative of equation of the rubber, E, and the gradient at $P$.
By the identity $\sin(\arctan(x)) = \frac{x}{\sqrt{x^2 + 1}}$,
$$F_P = T \frac{E'}{\sqrt{E'^2 + 1}} $$
But since the extension itself depends on $F$, we can get a differential equation:
$$E = \frac{T}{k} \frac{E'}{\sqrt{E'^2+1}} $$
I'd love to be able to solve this myself, and if someone could show me I'd be very grateful, but plugging this into Wolfram Alpha gives the solutions:
$$c+x=\sqrt{l^{2}-y^{2}}-l\operatorname{artanh}\left(\frac{\sqrt{l^{2}-y^{2}}}{l}\right)$$
and
$$c-x=\sqrt{l^{2}-y^{2}}-l\operatorname{artanh}\left(\frac{\sqrt{l^{2}-y^{2}}}{l}\right)$$
where $l = \frac{T}{k}$
Graphing these in desmos gives two curves which, when they intersect, seem to give a roughly correct-looking curve.
My questions are:
How can the differential equation be solved analytically?
What initial conditions am I missing to be able to eliminate $c$ and obtain a particular solution?
How do I relate $T$ to the weight of the ball in the middle, and does this affect the differential equation or provide us with initial conditions?