Proving that a sphere has a minimal surface to volume ratio using Calculus of Variations

Question

I know the problem is traditionally solved via the isoperimetric inequality, but I was hoping to solve it by minimizing a surface of revolution subject to a volume constraint.

The surface area of a surface of revolution is:

$$A=2\pi\int_{-1}^{1}y\sqrt{1+\dot{y}^2}~dx$$

and the volume will be:

$$V=\pi\int_{-1}^{1}y^2~dx$$

I'd like to show that the sphere (or in this case the function $x^2+y^2=1$) minimizes the surface area functional for any fixed volume.

The combined action will then be:

$$S=\int_{-1}^{1}2\pi y\sqrt{1+\dot{y}^2}+\lambda\pi y^2 dx$$

The Euler Lagrange Equation simplifies to:

$$-\frac{2 \pi y'(x)^2}{\sqrt{y'(x)^2+1}}+2 \pi \sqrt{y'(x)^2+1}+\frac{2 \pi y(x) y'(x)^2 y''(x)}{\left(y'(x)^2+1\right)^{3/2}}-\frac{2 \pi y(x) y''(x)}{\sqrt{y'(x)^2+1}}+2 \pi \lambda y(x)=0$$

I have no clear idea how to proceed from here. I can solve for $\lambda$ using Mathematica, and get

$$\lambda=\frac{y(x) y''(x)-y'(x)^2-1}{y(x) \left(y'(x)^2+1\right)^{3/2}}$$

The ODE solution is extremely convoluted, involving about half a dozen elliptic integrals, and doesn't seem to be going anywhere. Is there something fundamentally wrong with my approach, or is there an obvious next step in the problem?

EDIT: Using the Beltrami identity instead, and reversing the problem so the volume is varied and a constant surface area is the constraint, the EL equation can be simplified to:

$$y(x)^2+\frac{2\lambda y(x)}{\sqrt{y'(x)^2+1}}=k$$

Which, while much simpler, doesn't get me closer to an answer.

If $k$ is equal to $0$ and $y(0)$ is likewise, then the ODE reduces nicely, and can be solved. One solutions is:

$$y(x)=\sqrt{4 \lambda ^2-x^2}$$

Which is the equation of a circle (which when rotated around the axis becomes a sphere). However, some assumptions are made there which are not necessarily true.

Answer 1

I spent a few more hours on the problem, and eventually found the solution. No major breakthroughs, but a lot of algebra.

First, I reversed the problem, instead maximizing the volume subject to a surface area constraint. Not strictly necessary, but it simplifies the computation. I also gave the surface area a magnitude $k$, per user7530's comment. The new action is:

$$S=\int_{-\lambda/2}^{\lambda/2}\pi y^2 + 2\lambda\pi y\sqrt{1+\dot{y}^2}-\lambda k~dx$$

The integration boundaries were also changed for convenience.

Then, because the Lagrangian does not depend explicitly on the variable $x$, I was able to apply the Beltrami identity instead of the traditional EL equation.

$$\frac{d}{dx}\left(\pi y(x)^2 + 2\lambda\pi y(x) \sqrt{1+y'(x)^2}-\lambda k - \frac{2\pi\lambda y'(x)^2y(x)}{\sqrt{1+y'(x)^2}}\right) = 0$$

Integrating both sides, we find that the functional inside the derivative is equal to a constant. Because $y$ is equal to $0$ at $-\lambda/2$ and $\lambda/2$, and each term in the derivative depends on $y(x)$ except for the $\lambda k$ term, the arbitrary additive constant must be $-\lambda k$, which cancels with the $-\lambda k$ on the left side.

The DE becomes:

$$\pi y(x)^2 + 2\lambda\pi y(x) \sqrt{1+y'(x)^2} - \frac{2\pi\lambda y'(x)^2y(x)}{\sqrt{1+y'(x)^2}} = 0$$

Multiplying the second term by $\frac{\sqrt{1+y'(x)^2}}{\sqrt{1+y'(x)^2}}$, that in turn simplifies to:

$$y(x)^2 + \frac{2\lambda y(x)}{\sqrt{1+y'(x)^2}}=0$$

This can be solved for $y(x)$, and with the constant chosen to equal zero, the solution is the equation of a circle of radius $2\lambda$.

$$y(x)=\pm\sqrt{4 \lambda ^2-x^2}$$

Answer 2

There are a few problems with the formulation:

1. The volume constraint: right now you are constraining the total volume to be zero (check by differentiating $S$ with respect to $\lambda$). If you want the total volume to be equal to some fixed value $C$, you need instead the action

$$S = \int_{-1}^1 \left[2\pi y \sqrt{1+y'^2} + \lambda(\pi y^2-C/2)\right]\,dx.$$

2. The boundary conditions: your problem restricts your surface of revolution to touch the bounding points $(\pm 1,0,0)$. Of course, depending on the value of $C$ the sphere of volume $C$ will not usually touch these points. You can still minimize surface area, given these boundary conditions, but realize you are solving a harder version of the isoperimetric problem. (Which I also cannot solve at the moment).

Answer 3

I have to solve a similar problem, here are my thoughts.

We could express the problem in spherical coordinates with the domain $\Omega \in \mathbb{R}^3$ delimited by $\partial \Omega$, the Surface $A = \vert \partial \Omega \vert$ and the Volume $V = \vert \Omega \vert = constant$:

$$min\{A[R]\} = \int_{\partial\Omega} R^2(\varphi,\theta) {\rm sin}(\theta){\rm d}\varphi {\rm d}\theta$$ $$constant = V = \int_{\Omega} r^2 {\rm sin}(\theta) {\rm d} r {\rm d} \varphi {\rm d} \theta = \int_{\partial\Omega} \left( \int \limits_{0}^{R(\varphi,\theta)} r^2 {\rm d} r \right) {\rm sin}(\theta) {\rm d}\varphi {\rm d}\theta = \int_{\partial\Omega} \frac{R^3(\varphi,\theta)}{3} {\rm sin}(\theta) {\rm d}\varphi {\rm d}\theta $$

Now we have two integrals over the same domain $\partial\Omega$ and we can use formulas for problems under isoperimetric conditions.

With $F := R^2(\varphi,\theta){\rm sin}(\theta)$ and $G := \frac{R^3(\varphi,\theta)}{3} {\rm sin}(\theta)$ and $F^* = F + \lambda G$, we can use the formula $$ 0 = \underbrace{ \frac{{\rm d}}{{\rm d} \theta} \frac{ \partial F^* }{ \partial \frac{ \partial R }{ \partial \theta}} }_{=0} + \underbrace{ \frac{{\rm d}}{{\rm d} \varphi} \frac{ \partial F^* }{ \partial \frac{ \partial R }{ \partial \varphi}} }_{=0} - \frac{ \partial F^* }{ \partial R} $$

Therein is $$ F^* = \underbrace{ R^2(\varphi,\theta) {\rm sin}(\theta) }_{=F} + \lambda \cdot \underbrace{ \frac{R^3(\varphi,\theta)}{3} {\rm sin}(\theta) }_{=G} $$

So it follows $$ 0 = \frac{ \partial F^* }{ \partial R} = 2 R (\varphi,\theta){\rm sin}(\theta) + \lambda \cdot R^2(\varphi,\theta) {\rm sin}(\theta) $$

Solving the equation for $R$: $$ R(\varphi,\theta) = \frac{2}{\lambda} $$

So we have shown, that $R(\varphi,\theta) = constant$, because $\lambda$ is constant in $\varphi$ and $\theta$.

Answer 4

In the original notation, I decided to maximize the volume keeping the surface area fixed, the action-integral being: $$ I = \int_{-1}^{1} (πy^2 - \lambda 2πy\sqrt {1+y'^2} ) dx $$ Then, solving straight the Euler-Lagrange DE, I could reduce it to a Bernoulli type DE. Solving it, I got the result: $$ y'(x)= \frac {\sqrt{(2\lambda (y-C)-y^2)(2\lambda (y+C)+y^2)}}{y^2+2C\lambda} $$ Where $C$ is the constant of integration obtained in the aforementioned Bernoulli DE and $\lambda$ the Lagrange multiplier. For $C=0$ this solves to a circle, i.e. a spherical surface of revolution. However, I'm stuck at the case $C≠0$ : either this means there exists some other surface of revolution that can maximise the volume, or there is some way to prove, by ab initio methods, that $C$ must be zero. I humbly invite some suggestions over this.

Edit

With little algebraic manipulation we can show that $C< \frac {\lambda}{2}$. With $\lambda=1$ let $C=1/2n$ for a natural number $n$. The integral is then: $$ x+c=\int\frac{y^2+\frac{1}{n}}{\sqrt {4(1-\frac{1}{n})-(y^2-(2-\frac{1}{n}))^2}} $$ As $n \to \inf $ we get the sphere solution. What is the general solution $y_n$?