Symmetry of Solution to Classical 3-Dimensional Isoperimetric Problem

A while ago I attempted to solve the classical isoperimetric problem in 3-dimensions, namely "Find the surface that has the smallest surface area for a given volume".

At that time for me to write down the full euler-lagrange equations for such a problem was too tough, so I made the assumption that the solution to this problem was a surface of revolution about an axis.

The functional in question then becomes only a function of $y$ (from the calculus version of Pappus's Centroid Theorem).

Questions:

1) How is such an assumption justified? I remember reading through some results of Antonio Ros, Manuel Ritoré, Fred Almgren, Michael Hutchings et. al on the double bubble conjecture. I didn't really understand them, and I don't even remember the paper I looked at that had at least a useful result within my reach.

2) Upon substituting such a functional into the euler-lagrange equations (for the case of a functional dependent only on $y$ and $y'$), one gets the differential equation $\frac{2}{\sqrt{1+y'^2}} + \lambda y = C$, where $C$ is a constant and $\lambda$ the lagrange multiplier.

Now if $C = 0$ the equation of a circle is a solution, and one gets $C= 0$ by appplying the boundary conditions $y(a) = y(b)=0$, namely that the endpoints of such a curve (well curve because we are talking of a surface of revolution) lie on the $x-axis$.

What happens if $C$ is not zero? Apparently this would give rise to a different surface (as Delaunay) studied. Are there several solutions to the given differential equation that satisfy the problem?

Thanks.

-
Please at least use the minimum of thought when tagging questions: there already exists a (calculus-of-variations) tag, why did you create a (variational-calculus) tag? – Willie Wong Aug 3 '11 at 19:44
@Willie Wong There is something wrong with my browser. I tried typing tags like "linear-algebra" and it would not even appear. I just made a guess that that was the tag already created. – user38268 Aug 3 '11 at 20:39
Sounds like a bug. Please file a bug report at Meta.Math.StackExchange, taking care to include the operating system information, and the browser version information in the report. – Willie Wong Aug 3 '11 at 20:44
I think it was one of those things that just happened "then". I have been on this forum for about 3 months and never encountered something like this. Anyway for the question above do you have any thoughts to say about it? – user38268 Aug 3 '11 at 20:50
For (1): generally the method is to show that if there exists an extremizer such that it is not axisymmetric, you can perform a modification of the surface such that it becomes axisymmetric while reducing the surface area and/or increasing the volume content. In the present case: assume your surface is $C^1$. Try: Fix an axis. Define a new surface such that on each slice perpendicular to the axis, your replace the cross section by a circle of the same area. To show that the surface area is non-increasing under this will require some work. – Willie Wong Aug 3 '11 at 21:01

You could justify your reduction in the following way:

• For each level $\Gamma_{z_0}=\{(x,y,z) \in \Omega: z=z_0\}$ of your body $\Omega$ you can apply the isoperimetric inequality in two dimensions and conclude that the rearranged in the shape of a circle centered on the $z$-axis with the same area as $\Gamma_{z_0}$ has lower perimeter.

• Note that the new body which is now rotationally symmetric by the $z$-axis has the same volume as the initial body by using Cavallieri's principle.

• Using the coarea formula you can split the surface as an integral over the interval $[0,h]$ (you can assume that the body is contained in the strip $\{ 0 \leq z \leq h\}$) of the perimeters of the level sets. Since the level sets of the new body all have lower perimeters than the level sets of the initial body, this procedure reduces surface area.

-

The easiest proof of the isoperimetric problem that I have seen is the following one:

• take a candidate for the optimal solution
• take a plane such that the surface is splitted in half
• if the volume of the right part is bigger than the left one, you can build a more efficient surface by taking the right part and his symmetric with respect to the chosen plane. The total surface remains the same (you have previously splitted in half the surface) but the enclosed volume increase.

Now a surface that it is symmetric for each possible plane direction is the sphere. I am not sure if this proof works also for showing that this is the only possible surface.

Your problem now, I do not know if a similar proof exist for show that every optimal shape should be generated by revolution around an axis, but if it exist you can follow from "the optimal shape should be generated by revolution around an axis" that "the optimal shape should be generable by revolution around any axis" that it is the proof that the sphere is a solution.

-

Comment to the question (v3): Since OP mentions the Euler-Lagrange equations, it seems relevant to mention the following elementary derivation of the 3-dimensional isoperimetric inequality $$\tag{1} 36\pi V^2 \leq A^3,$$ although it additional assumes that the region is star-shaped.

Sketched proof: We can use spherical coordinates $(r,\theta,\varphi)$. Let $\chi\equiv \cos\theta$. The volume

$$\tag{2} V[r]~:=~ \frac{1}{3}\int_{-1}^{1}\!\mathrm{d}\chi \int_{0}^{2\pi}\!\mathrm{d}\varphi ~r(\chi,\varphi)^3$$

and surface area

$$\tag{3} A[r]~:=~ \int_{-1}^{1}\!\mathrm{d}\chi \int_{0}^{2\pi}\!\mathrm{d}\varphi ~r(\chi,\varphi)^2$$

are functionals of the surface profile function $r: [-1,1]\times [0,2\pi[\to [0,\infty]$.

Next use the method of Lagrange multipliers. We would like to maximize the volume $V[r]$ for fixed surface area $A[r]=A_0$. We therefore consider the functional

$$\tag{4} E[r]~:=~V[r]+\lambda (A_0-A[r]).$$

$$\tag{5} 0~=~\frac{\delta E[r]}{\delta r}~=~\frac{\delta V[r]}{\delta r}-\lambda\frac{\delta A[r]}{\delta r}~=~r(\chi,\varphi)^2- 2\lambda r(\chi,\varphi),$$
which leads to a stationary surface profile that is a constant independent of $\chi$ and $\phi$,
$$\tag{6} r(\chi,\varphi) ~=~ 2\lambda,$$
i.e. a ball. Together with the fact that the boundary profiles $r(\chi,\varphi)=0$ and $r(\chi,\varphi)=\infty$ are not relevant, this shows that balls maximize the volume (of a star-shaped region) for given fixed surface area.