Sign up ×
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.

The isoperimetric inequality on the sphere of radius 1 asserts that for any closed curve on the sphere, $$L^2 \geq A(4\pi - A)$$ where $L$ is the length of the curve and $A$ is the area it encloses. There are a number of proofs of this; I am looking for a proof using the calculus of variations in the spirit of the proof of the standard isoperimetric inequality on the plane. I don't think this can be done, but I thought I'd see if others have an idea.

First, let us review the variational proof of the isoperimetric inequality on the plane. By Green's theorem, the area enclosed by a curve $(x(t), y(t))$ is given by: $$A = \int_a^b x(t)y'(t)\, dt$$ On the other hand the length is given by: $$L = \int_a^b \sqrt{x'(t)^2 + y'(t)^2}\, dt$$ Using Lagrange multipliers, we can maximize $A$ while treating $L$ as a constraint, and the isoperimetric inequality follows.

To adapt this idea to the sphere we would need to use Stokes' theorem instead of Green's theorem to manufacture a line integral which calculates surface area. Namely, we would need a vector field $\textbf{F}$ on $\mathbb{R}^3$ such that if $C$ is a closed curve on the sphere which is the boundary of a piece $S$ of the sphere then: $$\int_C \textbf{F}\cdot \textbf{T}\, ds = \iint_S \text{curl}\, \textbf{F} \cdot \textbf{n}\, dS = \iint_S dS$$ In other words we want to choose $\textbf{F}$ so that $\text{curl}\, \textbf{F} = \textbf{n}$, the outward unit normal vector of the sphere. But there can be no vector field in $\mathbb{R}^3$ whose curl is $\textbf{n}$ since the integral of $\textbf{n}$ over the sphere is $4\pi$ while the integral of the curl of a vector field over any closed surface is $0$ by Stokes' theorem.

This seems to suggest that you can't use the calculus of variations to prove the isoperimetric inequality on any closed surface, which is a bit disappointing. Is there any way to get around this problem?

share|cite|improve this question

2 Answers 2

up vote 2 down vote accepted

You might want to look at the Gauss-Bonnet theorem, which, on the sphere, expresses the area enclosed by a curve in terms of the integral of curvature along the curve, and the exterior angles at any "corners" on the curve. To apply this theorem, however, requires that the curve be piecewise nice (as does your application of Green's theorem in the plane), so it really only proves the inequality for piecewise-nice (i.e., differentiable except at finitely many points) curves. That may be enough to satisfy you, in which case Gauss-Bonnet will be your friend for the sphere example.

(The GB theorem actually relates curvature along arcs, "curvature at corners" (i.e., exterior angles) and curvature inside the curve, but for the unit sphere, the total curvature on the area inside the curve is just the integral of 1 over that area, hence is equal to the area.)

In your application, the integral along the curve would be an integral of the curve's curvature (plus any exterior angles, if you want to allow broken curves); the constraint would be just like your $L$ (except that it'd have $x, y,$ and $z$ terms) and there'd be another constraint: $x^2(t) + y^2(t) + z^2(t) = 1$.

share|cite|improve this answer
I got this to work out - thanks! – Paul Siegel Apr 28 '14 at 6:18

While not directly answering your question I'd suggest you have a look at 'Isaac Chavel: Riemannian Geometry, A modern introduction' chapter V section 4 and section 5. There you will find an extensive study of isoperimetric inequalities on surfaces, including surfaces of positive curvature with a curvature bound from above.

If you are curious also the higher dimensional case is discussed in detail in that book (in later chapters, though...)

(I'm not sure what exactly you mean by a variational approach, the calculations in that book qualify definitely as variational...)

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.