Sphere and Isovolume problem

**TwistedOne151** · December 5th 2007, 12:16 PM

I'm trying to prove that the space region that contains maximum volume for a given fixed surface area is a sphere. I've done this for the analogous 2d isoperimetric problem using calculus of variations.
Here's what I have:
Let our region of space be V and the enclosing surface $\partial V$ . We want to minimize the volume $I = \iiint\limits_{V}\,dV$
subject to the constraint $S = \iint\limits_{\partial V} \, dS$ .

We parametrize $\partial V$ with parameters u and v. Then $dS=\left | \mathbf{r}_u \times \mathbf{r}_v \right | du dv$ , where $\mathbf{r}_u = \left (\frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}, \frac{\partial z}{\partial u} \right ) = \left (x_u , y_u , z_u \right )$ and $\mathbf{r}_v = \left (\frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}, \frac{\partial z}{\partial v} \right ) = \left (x_v , y_v , z_v \right )$ .
Let us define $\mathbf{F}(x,y,z) = \frac{x}{3}\mathbf{i} + \frac{y}{3}\mathbf{j} + \frac{z}{3}\mathbf{k}$ . Then $\nabla \cdot <br /> \mathbf{F}(x,y,z) = 1$ , and $\iiint\limits_{V}\nabla \cdot \mathbf{F}(x,y,z)\,dV = \iiint\limits_{V}\,dV = I$ .

Thus, by the divergence theorem,
$I = \iiint\limits_{V} \nabla \cdot \mathbf{F}(x,y,z) \,dV$
$= \iint\limits_{\partial V} \mathbf{F} \cdot \mathbf{\hat n} \, dS$ ,
where $\mathbf{\hat n}$ is the outward unit normal vector. For our parametrization of $\partial V$ , we have $\mathbf{\hat n} dS = \mathbf{r}_u \times \mathbf{r}_v du dv$ and thus
$I = \iint\limits_{\partial V} \mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v) \, du \, dv$ , and thus we want to find the extremum of:

$M = \iint\limits_{\partial V} \mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v) + \lambda \left | \mathbf{r}_u \times \mathbf{r}_v \right | \, du \, dV$

With $\mathbf{r}_u \times \mathbf{r}_v = (y_u z_v - z_u y_v) \mathbf{i} + (z_u x_v - x_u z_v) \mathbf{j} + (x_u y_v - y_u x_v) \mathbf{k}$

And so $\mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v) = \tfrac{1}{3} (y_u z_v - z_u y_v) x + (z_u x_v - x_u z_v) y + (x_u y_v - y_u x_v) z)$

and $| \mathbf{r}_u \times \mathbf{r}_v | = \sqrt{(y_u z_v - z_u y_v)^2 + (z_u x_v - x_u z_v)^2 + (x_u y_v - y_u x_v)^2}$

And we put these into the formula for M to get a double integral over dudv of a function $L(x,y,z,x_u,y_u,z_u,x_v,y_v,z_v)$ .

Were there only one independent variable, I could find the Euler-Lagrange equations for x, y and z from L; or more specifically, I'd use the lack of explicit dependence on the independent variable to apply the Beltrami Identity to find lower order differential equations for x, y, z. However, this has two independent variables. How do I set up and solve the Euler-Lagrange equations for this sort of problem? Is there a Beltrami Identity equivalent for multiple independent variables when the integrand has no explicit dependence on both of the independent variables?

--Kevin C.

**ThePerfectHacker** · December 5th 2007, 06:10 PM

Do you know that the 2 dimensional isoperimetric inequality can be established with Fourier analysis? But I cannot see how we can generalize that approach to 3 dimensions.

**Rebesques** · December 7th 2007, 05:40 AM

How do I set up and solve the Euler-Lagrange equations for this sort of problem?

There's more than one points to consider in that question. Even if you manage to set up the Euler-Lagrange eqns, I bet my measly paycheck there's no way on earth we can establish the full range of solutions. It's true, we can verify that spheres are solutions, but there will be no way to show there are no other surfaces to do the job.

I think there's a beter way to approach this. First, relax the conditions so we can work locally, and then show spheres are maximizers.

To this end, suppose we want to maximize the volume under the graph of a surface with fixed surface area, say over the unit disk in $\mathbb{R}^2$ . (Since all surfaces are locally graphs, this is not too demanding). Also let all those surfaces have zero values on $\partial D$ (which we can do without, but makes life easier). For nonnegative functions $u<img src=$ \rightarrow\mathbb{R}, \ u\in C^1_0" alt="u

\rightarrow\mathbb{R}, \ u\in C^1_0" /> we wish to minimize

$I(u)=\int\int\int_{V}dxdydu=\int\int_D u(x,y)dxdy$

along with the condition the sufrace area is fixed:

$S(u)=\int\int_D \sqrt{1+u_x^2+u^2_y}dxdy=S_0$ .

In this configuration, we have the major advantage that $I$ is linear, so it will attain maxima where $S$ does (truly, if $u$ is a minimizer, there is a Langrange multiplier $\lambda$ such that the Frechet derivatives satisfy $dI(u)+\lambda dS(u)=0\Rightarrow dS(u)=\lambda^{-1} I(u)$ ).

Now the proof follows from the classical isoperimetric inequality: $S$ is the surface area functional, and is maximized on spheres. Specially, the sphere with area $S_0$ is the unique maximizer in the considered class.

Now, we have showed the maximizers are locally spheres. But since spheres are isotropic, the global maximizer is but a sphere itself.

**mustkara** · February 20th 2008, 07:30 AM

I used the textbook

Isoperimetric Inequalities : Differential Geometric and Analytic Perspectives
Rainyland(Cocomartini) BookStore -University Textbooks Campus Reference Books

to study Isoperimetric. The discount bookstore offer attractive price.

Wish some help.

^_^

**Rebesques** · February 23rd 2008, 01:05 PM

Looks kind of hard for a textbook.

Anyway. Maybe someone will enjoy proving the following:

a) Among all convex plane sets of fixed diameter (*) the circle has maximum area.

b) Among all convex solids of fixed diameter, the sphere has maximum volume.

Enjoy

(*) The diameter of a set S is defined as d(S)=sup{|x-y| : x,y in S} (see? there's life after latex!).

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Sphere and Isovolume problem

Good reference Isoperimetric textbook

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