# Prove that the sphere is the only closed surface in $\mathbb{R}^3$ that minimizes the surface area to volume ratio.

It is well known that a sphere minimizes the surface area to volume ratio since it reaches equality in the Isoperimetric Inequality. I'm trying to prove that no other closed surface has this property.

Since the inequality says that $$36\pi V^2 \leq A^3$$ we can state the problem as:

Let $$\mathcal{S}$$ be a simple closed surface in $$\mathbb{R}^3$$ with surface area $$4\pi$$. Prove that if its volume is $$\dfrac{4}{3}\pi$$ then $$\mathcal{S} = \mathbf{S}^2$$.

My hypothesis:

If there is a surface $$\mathcal{S}\neq \mathbf{S}^2$$ but with the same surface area to volume ratio then there is a function $$F(t,x):I\times \mathbf{S}^2\to \mathbb{R}^3$$ so that:

$$F(t,x_0) \text{ is continuous }\forall x_0\in\mathbf{S}^2$$ $$F(t_0,\mathbf{S}^2) \text{ is a simple closed surface }\forall t_0\in I$$ $$F(0,\mathbf{S}^2) = \mathbf{S}^2\qquad F(1,\mathbf{S}^2) = \mathcal{S}$$ $$F(0,\mathbf{S}^2) \cong F(t,\mathbf{S}^2)\qquad \forall t\in I$$ $$V(F(t,\mathbf{S}^2)) = \frac{4}{3}\pi\qquad A(F(t,\mathbf{S}^2)) = 4\pi\qquad \forall t\in I$$

Where $$\cong$$ refers to equivalence via a bijective function, and $$V(X)$$ and $$A(X)$$ are, respectively, the volume and surface area functions.

In other words, there is some kind of continuous homeomorphism between both surfaces that conserves both volume and surface area.

Then by using variation calculus we can prove that if a surface is "almost" a sphere, it has a higher $$S/V$$, thus there cannot be such function $$F$$ and therefore there is no such surface $$\mathcal{S}$$.

Perhaps my hypothesis is false. Feel free to prove the statement without using it.

The problem of to show that the surface with minimal area that enclose a given volume is a sphere is not at all simple.

Using the calculus of variation we can show that the surface must have constant curvature. I sketch the proof:

Given a smooth closed surface $S$ in $\mathbb{R}^3$, let $A$ its area and $V$ the volume enclosed. As varied surface consider the surface $S(t)$ obtained by displacing each point of $S$ by a vector $th\vec n$, where $\vec n$ is the normal to the surface and $h$ is a smooth real valued function $h:S\rightarrow \mathbb{R}$. The first variations of the area and the volume are:

$$A'(0)=-\int_ShHdA \qquad V'(0)=\int_ShdA$$

where $H$ is the mean curvature of the surface with respect to the normal.

Now, to have the minimum area for given volume, it must be true that whenever $V'=0$ also $A'=0$, so $H$ must be a constant on $S$.

But now the question is to show that a surface of constant mean curvature is necessarily a sphere and this can be proved, but the proof is not simple. A first proof was given by Liebmann in 1900 for compact strictly convex surfaces, generalized by Aleksandrov in 1962 to surfaces with some kind of self-intersection. But, as far as i know, the question of the possible existence of surfaces of any topological type with self intersection and constant mean curvature, is unanswered.

For a complete treatment of this topic you can see here.

• A self-intersecting surface of constant curvature does exist. Consider a torus. Usually the major radius R > minor radius r. When R < r, the torus self intersects. When R is small, the torus resembles a sphere with 2 sheets. The sheets are connected near the north and south poles. When r = 0, the sheets connect only at the north and south poles, and the sheets lie on top of each other. Commented Nov 29, 2015 at 13:39
• Tori of revolution do not have constant mean curvature. Nevertheless there are cmc tori, for example the Wente torus is one such, and yes it does have self-intersections. Commented Nov 24, 2018 at 13:21

If I am not mistaking, the statement

Sphere is the only closed surface in $\Bbb R^3$ that minimizes the surface area to volume ratio.

is equivalent to either of the following statements:

1. Sphere is the only closed surface in $\Bbb R^3$ that minimizes surface area, enclosed volume fixed.
2. Sphere is the only closed surface in $\Bbb R^3$ that maximizes enclosed volume, surface area fixed.

I am going to try to sketch proof of the first statement.

## Statement:

Assume that $\mathcal S$ is 2D closed (Riemannian) manifold embedded in $\Bbb R^3$ with enclosed volume $V$. Then $\mathcal S$ has minimal possible surface area $\implies\mathcal S = S^2$.

### Proof (sketch):

1. Take an arbitrary compact Riemannian 2-manifold $\mathcal{M}$ of enclosed volume $V$ embedded in $\Bbb R^3$.
2. Recall that a surface is called minimal if it minimizes area.
• Minimal surfaces are the minimizers of mean curvature flow, i.e. they have constant mean curvature [1]. (This relation is supported by the First variation of area formula [2], which describes the relation between total surface area and its mean curvature).
3. Under the volume-presuming mean curvature flow any compact surface will evolve into a compact surface with constant mean curvature.
• The mean curvature of compact surface in $\Bbb R^3$ cannot be zero, because the maximum principle for harmonic functions implies that there are no compact complete surfaces of zero mean curvature in $\Bbb R^3$.
4. The only surface in $\Bbb R^3$ with constant nonzero curvature is sphere $S^2$ [2].
• Closed surfaces in $\Bbb R^3$ cannot have zero mean curvature, therefore the only surface minimizing its area for a fixed volume is $S^2$.

In fact, these conclusions are supported by the formal derivations in [1, sec. 9-10], where it is proved that under volume-presuming mean curvature flow any compact surface will evolve into sphere.

In relation to your proof, you can argue that if the surface $\mathcal S$ has the same volume and surface area as $S^2$, then it must also have the same (constant nonzero) mean curvature. Otherwise it would not be a minimizer of the volume-preserving mean curvature flow, i.e. it could have been deformed into something with the same volume but smaller surface area. But if it has the same constant nonzero mean curvature as $S^2$, then it must coincide with $S^2$ by Alexandrov's theorem them. Contradiction.

• Among DeLaunay surfaces there are no closed spaces? Commented May 28, 2015 at 15:35
• Aren't these unbounded? That's the same as unduloids, is that correct?