Soap Film and Minimal Surface

soap_filmThe geometry of soap films and soap bubbles

If we dip two wire rings into a solution of soapy water, then what is the surface formed by the soap film ? This question is known as the minimal surface problem, or the Plateau problem. This post proposes to discuss a simple solution based on variational calculus and elementary computations.


Consider a soap film between two coaxial circular rings of the same radius R. We denote by z the mono-dimensional coordinate of the central axis. The circular rings are positioned at z=-L and z=+L respectively. By symmetry arguments, neglecting the gravity, the film corresponds to a surface of revolution along the z axis with radius r(z). Clearly, we have r(-L)=r(+L)=R. The length element along a meridian is ds^2=dz^2+dr^2, which corresponds to a displacement along both z and r coordinates.



The free energy of the soap film is E=\sigma S where \sigma is the film tension and S is the area of the film. At the thermodynamic equilibrium, the free energy is minimized. Thus, the minimization of the energy is equivalent to the minimization of the surface. The area element dS=2 \pi r ds generates the surface as the integral

\displaystyle S=\int 2 \pi r ds=2 \pi \int_{-L}^{+L}r \sqrt{1+r'^2}dz


In elementary calculus, the derivative of a function at an extremum is zero. Similarly, in variational calculus, the variation of a functional at an extremum is zero, namely

\displaystyle \delta S=S(r + \delta r) - S(r)=0

The quantity \delta r represents a small variation of the function r, which must satisfy the boundary conditions

\displaystyle \delta r(-L) = \delta r(+L) = 0

Note that the condition \delta S=0 is necessary but not sufficient to minimize the surface, since an extremum is not always a minimum. We compute

\displaystyle \delta S=2\pi\int_{-L}^{+L}\delta\left(r\sqrt{1+r'^{2}}\right)dz=0

\displaystyle \int_{-L}^{+L}\delta r\sqrt{1+r'^{2}}dz+\int_{-L}^{+L}r\delta\sqrt{1+r'^{2}}dz=0

\displaystyle \int_{-L}^{+L}\sqrt{1+r'^{2}}\delta r dz+\int_{-L}^{+L}\frac{r r'}{\sqrt{1+r'^{2}}}\delta r'dz=0

An integration by parts on the second integral allows to use \delta r instead of \delta r'

\displaystyle \int_{-L}^{+L}\frac{r r'}{\sqrt{1+r'^{2}}}\delta r'dz=\left[\frac{r r'}{\sqrt{1+r'^{2}}}\delta r\right]_{-L}^{+L}-\int_{-L}^{+L}\frac{d}{dz}\left(\frac{r r'}{\sqrt{1+r'^{2}}}\right)\delta rdz

The term in square bracket cancels due to the boundary conditions, hence

\displaystyle \int_{-L}^{+L}\sqrt{1+r'^{2}}\delta r dz-\int_{-L}^{+L}\frac{d}{dz}\left(\frac{r r'}{\sqrt{1+r'^{2}}}\right)\delta r dz=0

\displaystyle \int_{-L}^{+L}\left\{ \sqrt{1+r'^{2}}-\frac{d}{dz}\left(\frac{r r'}{\sqrt{1+r'^{2}}}\right)\right\} \delta r dz=0

The term in brace bracket cancels since \delta r is arbitrary, hence

\displaystyle \frac{d}{dz}\left(\frac{r r'}{\sqrt{1+r'^{2}}}\right)=\sqrt{1+r'^{2}}

The derivative is cumbersome but straightforward to compute, we arrive at the expression

\displaystyle \boxed{r r''=1+r'^2}


The differential equation can be solved in two steps of integration.

First, we proceed with basic calculus, using \left(1+r'^2\right)'=2 r' r''

\displaystyle \frac{r'}{r}=\frac{\left(1+r'^{2}\right)^{'}}{2\left(1+r'^{2}\right)}

\displaystyle \left(\log r\right)^{'}=\left(\log\sqrt{1+r'^{2}}\right)^{'}

\displaystyle \log r=C+\log\sqrt{1+r'^{2}}

\displaystyle r=K\sqrt{1+r'^{2}}

Second, we proceed with hyperbolic trigonometry

\displaystyle r'=\frac{dr}{dz}=\frac{\sqrt{r^2-K^2}}{K}

\displaystyle dz = K \frac{dr}{\sqrt{r^2-K^2}}

\displaystyle z = K \int\frac{dr}{\sqrt{r^2-K^2}}

\displaystyle z = K \cosh^{-1}\left(\frac{r}{K}\right) + \kappa

\displaystyle r = K \cosh\left(\frac{z - \kappa}{K}\right)

In particular, at z = \pm L we have

\displaystyle K \cosh\left(\frac{L - \kappa}{K}\right) = K \cosh\left(\frac{-L - \kappa}{K}\right)

Since \cosh is an even function we have either L - \kappa = -L - \kappa or L - \kappa = -\left(-L - \kappa\right). The first case would imply L = -L, which is inconsistent. The second case implies \kappa = 0. Thus, the final solution is

\displaystyle \boxed{r(z)=K\cosh\left(\frac{z}{K}\right)}


We analyze the solution when L=1. This implies R=K\cosh(1/K), which is plotted as follows


There is a minimum (positive) radius, which can be approximated with Mathematica

FindMinimum[K*Cosh[1/K], {K, 1}]
{1.50888, {K -> 0.833557}}

This means that the radius has a critical value \simeq 1.509. There are two solutions above this value, and no solution below. For example, if we choose R=2 then we solve the equation for K

NSolve[K*Cosh[1/K] == 2, K, Reals]
{{K -> 0.47019}, {K -> 1.69668}}

We obtain a stable catenoid solution for K=1.69668, which is a minimal surface


and an unstable catenoid solution for K=0.47019, which is not a minimal surface (the soap film will pop)



  1. Dierkes et al., Minimal Surfaces, Springer, 2010
  2. Osserman, A Survey of Minimal Surfaces, Dover Publications, 2014

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