# Equation satisfied by a rubber sheet

## Main Question or Discussion Point

Greetings,

How can I derive the equation satisfied by a rubber sheet? I know that it reduces to Laplace's equation when the sheet is quite planar. I found the expression for it on p.113 in Introduction to Electrodynamics by Griffiths. Thanks in advance.

Edit: I have attached the expression as jpg file.

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Simon Bridge
Homework Helper
Salutations;
The usual approach would be to consider a free-body diagram for a small element of the sheet.

Thanks for the answer. Can you elaborate more on that plese?

Simon Bridge
Homework Helper
It involves modelling the it as an array of small masses connected by springs.
The exact details depend on the geometry of your sheet and what you are doing with it - which you haven't said.

eg. a horizontal sheet of uniform thickness ##t## and density ##\rho## can be divided into small volume elements ##dV = tdxdy## each with mass ##dm=\rho dV##. The one between (x,y) and (x+dx, y+dy) will be acted on by gravity is the -z direction and a bunch of spring forces along the sheet surface - the spring constants being a material property of the rubber. The directions and magnitudes of the spring forces depends on the nearby distortions of the sheet.

The equilibrium shape of the sheet will be one where all the forces for every volume element sum to zero - given some boundary condition (i.e. what is holding the sheet up?) You'd want to find this by minimizing the action across the sheet (look it up).

The above example will produce an equation z=f(x,y) which has the familiar sag in the middle.

In general, displacements from equilibrium in the sheet will obey the 2D wave equation.

Thanks for the answer. I just saw this equation in Griffiths and wondered how he derived it. I do not know the underlying assumptions or the model.

Simon Bridge
Homework Helper
By inspection - it looks like Griffith has minimized the action using the general method outlined.
The text suggests that there was an earlier expression derived - probably simpler, using a simplified model.
Examining that should help - but authors will sometimes expose you to stuff like this just so you can get an idea of (a) how more there is to learn and (b) how close such a simple model (the previous one) actually is to a more complete treatment.

Septim
By inspection - it looks like Griffith has minimized the action using the general method outlined.
The text suggests that there was an earlier expression derived - probably simpler, using a simplified model.
Examining that should help - but authors will sometimes expose you to stuff like this just so you can get an idea of (a) how more there is to learn and (b) how close such a simple model (the previous one) actually is to a more complete treatment.
Dear Simon Bridge,

I am sorry to resurrect this thread, but I am still oblivious on how to proceed to get the equation satisfied by a rubber sheet from scratch. Okay I know the principle of action from classical mechanics, yet how can I apply it to the case of a rubber sheet?

Simon Bridge
Homework Helper
Have you tried looking for other examples?

Nidum
Gold Member
Look up membrane stress .

Septim
Have you tried looking for other examples?
By other examples you mean 1-D case or something? I could not find a rigorous derivation of the equation satisfied by a rubber sheet.

Look up membrane stress .
Do you have a particular source in mind?

Shouldn't you tell us what ## V ## is? At any rate the area of a curved surface can be calculated as ## A = \int dx (a) = \int dx \sqrt{1 + (\nabla h(x))^{2}} ## with ## h(x) ## being the heigth of the surface. Your equations looks like a ficks type of law with "diffusion" coefficient given by a, ala ## \nabla \cdot (\frac{1}{a} \nabla h(x))=0 ##. I dont know if that helps at all :|

Nidum
Gold Member
How can I derive the equation satisfied by a rubber sheet? I know that it reduces to Laplace's equation when the sheet is quite planar. I found the expression for it on p.113 in Introduction to Electrodynamics by Griffiths. Thanks in advance.
Could you please explain more clearly what you are actually trying to do ?

Shouldn't you tell us what ## V ## is? At any rate the area of a curved surface can be calculated as ## A = \int dx (a) = \int dx \sqrt{1 + (\nabla h(x))^{2}} ## with ## h(x) ## being the heigth of the surface. Your equations looks like a ficks type of law with "diffusion" coefficient given by a, ala ## \nabla \cdot (\frac{1}{a} \nabla h(x))=0 ##. I dont know if that helps at all :|
I am sorry for the inconvenience caused, V(x,y) denotes the height of the rubber sheet above the point (x,y) in this case. By the way do you say trying to minimize the area will work?

Could you please explain more clearly what you are actually trying to do ?
My aim is to drive the equation for the height V(x,y) of the rubber sheet as a function of the position (x,y) on the sheet. This equation is:
## \frac{\partial}{\partial x}(g \frac{\partial V}{\partial x}) + \frac{\partial}{\partial y}(g \frac{\partial V}{\partial y}) = 0 ##
This equation is a differential equation and when the membrane is more or less planar it reduces to Laplace's equation. The g is defined as follows:
## g = \frac{1}{\sqrt{1 + (\frac{\partial V}{\partial x})^2 + (\frac{\partial V}{\partial y})^2}} ##

I am sorry for the inconvenience caused, V(x,y) denotes the height of the rubber sheet above the point (x,y) in this case. By the way do you say trying to minimize the area will work?
Yes wouldn't that work? If ## A[h] = \int dx \sqrt{1 + (\nabla h)^{2} } ## doesn't ## \frac{\delta A}{\delta h} = 0## equal your equation?

Yes wouldn't that work? If ## A[h] = \int dx \sqrt{1 + (\nabla h)^{2} } ## doesn't ## \frac{\delta A}{\delta h} = 0## equal your equation?
Okay I should do the math and see I am a bit rusty on variational calculus, yet I will try. Do you think this will really give the answer?

Nidum
Gold Member
But what forces are causing the deflection ???

Lets just do it. Since ## A[h] ## is only dependent on the gradient of ## h ## the variational derivative reduces to
## \frac{\delta A}{\delta h} =- \nabla \cdot \frac{\partial \sqrt{1 + (\nabla h)^{2} } }{\partial \nabla h} =- \nabla \cdot \left( \frac{1}{\sqrt{1 + (\nabla h)^{2} } } \nabla h \right) ##. I might have ignored some important points about vectors and such but the result looks fairly good.

"It involves modelling the it as an array of small masses connected by springs."

Suppose we ignored gravity and just considered this infinitely thin sheet as modeled by a square lattice of masses, with masses that are adjacent in the x- and y-directions connected by springs. And then take a limit as the fineness of the lattice approaches 0.

Question: Will this model have the isotropy that a typical rubber sheet has?

Lets just do it. Since ## A[h] ## is only dependent on the gradient of ## h ## the variational derivative reduces to
## \frac{\delta A}{\delta h} =- \nabla \cdot \frac{\partial \sqrt{1 + (\nabla h)^{2} } }{\partial \nabla h} =- \nabla \cdot \left( \frac{1}{\sqrt{1 + (\nabla h)^{2} } } \nabla h \right) ##. I might have ignored some important points about vectors and such but the result looks fairly good.
Thanks for the answer actually I was not particularly sure how Euler-Lagrange equations worked for multiple integrals with more than one independent variable. As far as I understand the rubber membrane acts like a soap film. Therefore it tries to minimize its surface area for a given set of boundary conditions(say being stretched over a membrane or something. By the way I am also unsure about your notation; however the approach is totally correct and I was able to derive the exact equation with the same reasoning. I found some lecture notes about calculus of variations in multi-dimensional case with multiple independent variables on this site.

"It involves modelling the it as an array of small masses connected by springs."

Suppose we ignored gravity and just considered this infinitely thin sheet as modeled by a square lattice of masses, with masses that are adjacent in the x- and y-directions connected by springs. And then take a limit as the fineness of the lattice approaches 0.

Question: Will this model have the isotropy that a typical rubber sheet has?
I do not think it will have the same isotropy as the rubber sheet. By the way I was able to derive the equation using calculus of variations for multiple independent variables as suggested. I also corresponded privately with Prof. Griffiths. He sent two articles on waves in strings and told me that it is easy to move from the one dimensional case to the two dimensional case easily.

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Terminology: Surfaces that minimize the surface area inside any sufficiently small closed curve on the surface are called minimal surfaces (not minimum surfaces).

Terminology: Surfaces that minimize the surface area inside any sufficiently small closed curve on the surface are called minimal surfaces (not minimum surfaces).
Thanks. I'm finishing my post up now. Maybe you can check it for errors.

Hallo there,

I also saw the equation in Griffiths. After some searching I found a satisfactory explanation.

Here is a short derivation.
There is a more elaborate explanation in the attachment!

This equation is the Minimal surface equation derived by Lagrange in 1762 using the Euler-Lagrange equation.
One can also derive the equation directly by using calculus of variations to minimize surface area. The simplest derivation with this method I found here. (I have also done this in the attachment.)

Physical interpretation

Griffiths talks about an equation describing a stretched rubber sheet. Though, traditionally this equation is used to describe soap films. Namely, surface tension ##\sigma \doteq FL^-1 \doteq JL^-2## is in units of energy per area. As energy is minimized, the surface area of soap films is minimized.
Rubber approximates the equation if it is not to thick and shearing stresses are neglected. Namely, thick rubber will have additional resistance to bending and surface tension is only a "pulling" and not a "turning" force. If we assume the rubber is flat enough bending forces and twisting forces will be negligible.

Surface area equation

Firstly, we want to find an equation for the total surface area given some height function / surface function ##h(x,y)## on a domain ##x,y\in\Omega\subset\mathbb{R}## with a boundary ##\partial \Omega##. (You can think of the domain as a shadow directly below the surface and the boundary as the edge of that shadow.) A nice visualization and derivation you can find here.
Essentially, you chop up the the surface into area elements ##\Delta A## with a base of sides ##\Delta x## by ##\Delta y##. Then, the area is spanned by two vectors ##\Delta \textbf{u} = (\Delta x, 0, \frac{\partial f}{\partial x})^\top## and##\Delta \textbf{v} = (0, \Delta y, \frac{\partial f}{\partial x})^\top##. Thus,
$$\Delta A = \Delta |u\times v|= \sqrt{1+(\frac{\partial f}{\partial x})^2+(\frac{\partial f}{\partial y})^2}$$ and therefore the area of a function ##f(x,y)## over a domain ##x,y\in \Omega \subset \mathbb{R}## is given by $$A_\Omega (f) = \int_\Omega \sqrt{1+(\frac{\partial f}{\partial x})^2+(\frac{\partial f}{\partial y})^2} dxdy$$.

Minimizing functional by solving the Euler-Lagrange equation

The Euler-Lagrange equation for a function ##f(x,y)## with two variables that minimizes a functional of the form $$I[f] = \int_\Omega \mathcal{L}(x,y,f,\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})dxdy$$ is $$\frac{\partial \mathcal{L}}{\partial f} - \frac{d}{dx}(\frac{\partial \mathcal{L}}{\partial f_x})-\frac{d}{dy}(\frac{\partial \mathcal{L}}{\partial f_y})=0 \; \; | \; \; f_x = \frac{\partial f}{\partial x}, \; f_y = \frac{\partial f}{\partial y}$$

Identifying that the solution to the area functional ##A_\Omega## should be the optimal surface \ height function ##h(x,y)##.

Deriving minimum surface equation

Suppose we have some function ##h: \Omega \rightarrow \mathbb{R}^2## defined on a domain ##\Omega \subset \mathbb{R^2}## with boundary ##\partial \Omega## and this function minimizes the area functional ##A_\Omega (h)##. Then, for the set of all possible functions on this domain ##f:\Omega \rightarrow \mathbb{R}^2## with the same boundary conditions ##h(x_b,y_b)=f(x_b,y_b) \; \forall x_b,y_b\in \partial \Omega## we have $$\textrm{min}\; A_\Omega(f) = A_\Omega (h)$$.
Now, we identify ##\mathcal{L} = \sqrt{1+(\frac{\partial h}{\partial x})^2+(\frac{\partial h}{\partial y})^2}## must satisfy the Euler-Lagrange equation. As $$\frac{\partial \mathcal{L}}{\partial h} - \frac{d}{dx}(\frac{\partial \mathcal{L}}{\partial h_x})-\frac{d}{dy}(\frac{\partial \mathcal{L}}{\partial h_y})=0 \; \; | \; \; h_x = \frac{\partial h}{\partial x}, \; h_y = \frac{\partial h}{\partial y}$$ $$\frac{\partial \mathcal{L}}{\partial h}=0, \; \frac{\partial \mathcal{L}}{\partial h_x}=\frac{h_x}{\mathcal{L}}, \; \frac{\partial \mathcal{L}}{\partial h_y}=\frac{h_y}{\mathcal{L}}$$ we conclude $$\frac{\partial}{\partial x} \left( \frac{\frac{\partial h}{\partial x}}{\sqrt{1+(\frac{\partial h}{\partial x})^2+ (\frac{\partial h}{\partial y})^2}} \right) + \frac{\partial}{\partial y} \left( \frac{\frac{\partial h}{\partial y}}{\sqrt{1+(\frac{\partial h}{\partial x})^2+ (\frac{\partial h}{\partial y})^2}} \right) = 0$$.
or more compactly $$\nabla \cdot \left( \frac{\nabla h}{\sqrt{1+|\nabla h|^2}} \right) = 0$$ which is even generalizable for a gradient of any order, as explained in the attachment.

Two dimensional equivalent

We can similarly solve for the two dimensional case. Assume we have a function ##f(x,y)## in domain ##x,y\in \Omega \subset \mathbb(R)## with boundary ##(x_0,x_1) = \partial \Omega##. Now we will minimize the line functional $$L(f)=\int_{x_0}^{x_1} \sqrt{1+(\frac{\partial f}{\partial x})^2} dx$$.
Defining ##\textrm{min}\; L_\Omega(f) = L_\Omega (y)## we find $$\frac{\partial}{\partial x} \left( \frac{y_x}{\sqrt{1+(y_x)^2}} \right)=0$$ Now, we can solve this for the given boundary conditions $$\frac{y_x}{\sqrt{1+(y_x)^2})} = c$$ $$y_x=c\sqrt{1+(y_x)^2} \rightarrow (y_x)^2 =\frac{c^2}{1-c^2}$$ concluding $$y(x)=c_0 x+c_1 \rightarrow y(x) = (y(x_1)-y(x_0))x+x_0$$ The line functional / arc length is optimized for a straight line between the two boundary points.

Similarity to Laplace's equation

We can obtain Laplace's equation from the minimum surface equation by assuming the surface is almost planar. Namely, then ##(\frac{\partial h}{\partial x})^2 +(\frac{\partial h}{\partial y})^2\approx 0## and the equation will reduce to ##\nabla^2 h=0##

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