# 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.

• Hi, Welcome to Physics.S.E. Mar 27 '16 at 4:49
• (+1) The plastic surface will have two forces acting on it. The weight of water towards the ground and the elastic force due to its change of shape(shear). But how to write the e.o.m. is puzzling me.
– Ari
Mar 27 '16 at 4:58
• A heuristic observation: pretend your domain is circular, so that the problem reduces to 1d. The resting position of, eg, a chain (clamped at both ends) under the action of gravity is a catenary curve (which looks a like a parabola), and hence the 2d surface would be something like a paraboloid. This holds even when the mass density is variable. But I agree with you, to actually solve this, I'd start with a Lagrangian (but I have no idea what form it takes). Maybe the book by Love would be of use. Mar 27 '16 at 5:41
• This problem boils down to the eigenvalue problem for the Laplacian operator so it should produce a $J_0(r)$ solution, for small $r$ close to parabolic indeed. Mar 27 '16 at 7:16
• @MaximUmansky can you please explain your reasoning for this, or provide a link that I could read :)? Mar 27 '16 at 10:17

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.