Shape of water on top of a thin sheet of stretched plastic Consider a thin sheet of plastic (a square sheet for simplicity) that is stretched taught in a plane parallel to the ground. If a volume of water is then placed on top of the thin plastic sheet, then the water will further stretch the plastic and create a slight depression (assuming the plastic doesn't break). My question is: What shape will this volume of water be (or what shape is the bottom curved surface of the water) after it is allowed to settle? You can check out this video to see an example of what I mean https://www.youtube.com/watch?v=eeSyHgO5fmQ . The guy in the video says that it is almost a perfect paraboloid, but I don't see why it should be. This seems like a problem that can be solved using the calculus of variations, but I am stuck as to what the constraint/s should be. 
 A: Let's first recall the wave equation for a membrane 
$\partial_{tt} u = c^2 \Delta u$ where $u(x,y)$ is the vertical displacement. See, e.g., https://en.wikipedia.org/wiki/Vibrations_of_a_circular_membrane.
One can recognize the left-hand side $\partial_{tt} u$ as the acceleration of a membrane element and $\Delta u$ is the vertical force provided by a stretched membrane. A derivation of this is given, e.g., in 
http://www.math.iit.edu/~fass/Notes461_Ch7.pdf.
For a static problem we need only the force term which is the Laplacian.
The water lying on the membrane provides a vertical force on a membrane surface element proportional to the water column height above it which is the vertical displacement of the membrane $u(x,y)$ counted from the water surface level, $F = \rho_{water} g u(x,y) dx dy$. So equating the vertical forces on each surface element of the membrane one arrives at the eigenvalue equation
$\Delta u = \lambda u$. A solution with azimuthal symmetry $u(r)$, non-singular at $r=0$, is the Bessel function $J_0$. Of course all this is formally valid only for small displacements as this is all just linear theory.
