Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the formula for the planar curve through $(\pm a,0)$ of fixed length $l$ which has minimal-area surface of revolution when rotated about the x-axis?

I get the area of the surface to be $2\pi\int^a_{-a}y\sqrt{1+y'^2}dx$. Then the first integral of the Euler-Lagrange equation (Beltrami identity) is $y\sqrt{1+y'^2}-\frac{yy'^2}{\sqrt{1+y'^2}} = k$, which has general solution $y(x)=k\cosh(\frac{x-C}{k})$. But putting in the initial conditions $y(\pm a)=0$ gives no nontrivial solutions!

Sources such as here say the general solution is indeed $y(x)=k\cosh(\frac{x-C}{k})$, but only when the endpoints of the curve are not on the x-axis. The question I'm looking at says the solution in this case is $y(x)=k(\cosh\frac{x}{k}-\cosh\frac{A}{k})$, but this doesn't seem to satisfy the right ODE!

Many thanks for any help with this!

share|cite|improve this question
Could you please make your question a bit more comprehensible? What role does $l$ play, you mention it only once. The $\cosh$ curve have infinite length. What is $A$? – user20266 Jul 9 '12 at 17:02

The only minimal surfaces of revolution about the $x$-axis are a flat plane (orthogonal to the axis) and the catenoid. The plane does intersect the axis, the catenoid simply does not. That's life. The catenoid can be translated and scaled without loosing minimality. Meanwhile, it is physically stable only for a bounded subset. If we take two equal circles of wire, and dip them in soap solution so as to form a sort of hollow tube of soap film between them, the circles (if kept co-axial) can be separated only to roughly 2/3 of the diameter of the circles. More separation and the soap film pops.

To sum up, if you have a problem in the calculus of variations, there is little assurance that a smooth/continuous minimizer exists. If you attempt to work with this with just O.D.E. techniques the result is often nonsense.

share|cite|improve this answer
So, does this mean there's a sequence of smooth surfaces with area tending toward a non-zero limit, but only a non-smooth surface achieves the limit? or does it mean there's a sequence of smooth surfaces with area tending toward zero? – Gerry Myerson Jul 10 '12 at 1:01
@Gerry, for two coaxial circles in parallel planes, if they are too far apart the only minimal surface with that as boundary is a pair of disks. There is a lower bound (which I do not know offhand) for a connected annular surface, smooth, that does have the two circles as boundary. There is an easy argument using the maximum principle for elliptic PDE that shows that the circles cannot be separated more than their diameter and have a spanning connected surface-make a second copy, rotate it 90 degrees, pull them apart and push back together until first contact. This creates an internal – Will Jagy Jul 10 '12 at 1:31
@Gerry contact. With minimal surfaces, tangency with one surface locally on one side of the other is prohibited. Tangency when both pass through each other arbitrarily close to the point of tangency is fine, compare the $xy$-plane and the graph of $z = \mbox{Real}(x + i y)^3$ near the origin. In fact locally harmonic function graphs are quite a good approximation to minimal surfaces, as the Laplacian is the highest order part of the minimal surface operator. If $z = f(x,y)$ gives a minimal surface as graph, $\nabla f(0,0) = \vec{0},$ then $\Delta f(0,0) = 0.$ – Will Jagy Jul 10 '12 at 1:37
@Gerry, I think you are right about the non-smooth surface realizing the minimum. Take a smooth disk on each circle with a central opening, join by a smooth very narrow tube between them. This area is arbitrarily close to the area of the two discs. Make the tube thinner and thinner, it disappears. So I withdraw the comment that there is some reasonable minimum, there is not. – Will Jagy Jul 10 '12 at 1:42
I was thinking about the original problem, where the endpoints of the curve are on the axis, and the length of the curve is fixed. – Gerry Myerson Jul 10 '12 at 1:46
up vote 1 down vote accepted

Answering my own question here...

The fact that curve length is fixed amounts to a functional constraint. So we need to use both the Euler-Lagrange equation and Lagrange multipliers. The functional to minimise is $\int^a_{-a}(2\pi y-\lambda)\sqrt{1+y'^2}dx$, where $\lambda$ is a Lagrange multiplier. For this functional, the first integral of the Euler-Lagrange equation gives a different ODE from what I got above, which has the solution $y(x)=k(\cosh\frac{x}{k}-\cosh\frac{a}{k})$ for some $k$ depending on $l$, as required.

@Will Jagy: given the endpoint constraints but not the length constraint, the curve which absolutely minimises surface area would be the x-axis, which gives a trivial surface. I can't see where you get the flat plane from? To get that as a surface of revolution, the curve would have to be a vertical line, but this doesn't satisfy the endpoint constraints!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.