Chaitanya's Random Pages

March 20, 2011

The Isoperimetric Inequality

Filed under: mathematics — ckrao @ 1:50 am

The following result has been known since Ancient Greek times if not earlier, yet it was not until the 19th century that it was rigorously proven.

Among all regions in the plane with fixed perimeter, the circle encloses the greatest area.

I recently came across a solution to a special case of the isoperimetric inequality (smooth curves) using Fourier series which I shall outline here. This is based largely on reference [1] with support from [2-5]. Let the plane curve C be parametrised in the complex plane by \displaystyle z:[0,2\pi]\rightarrow \mathbb{C}, \displaystyle z(t) \in \mathbb{C}. We assume the curve is smooth, meaning its derivative z'(t) exists and is continuous. Also assume the curve is simple (z(t) is one-one on [0, 2\pi)) and closed (so that z(0) = z(\pi)). The length of the curve is given by

\displaystyle P = \int_0^{2\pi} |z'(t)|\ dt

and the area is

\displaystyle A = \frac{1}{2i} \int_C \overline{z} \ dz = \frac{1}{2i} \int_0^{2\pi}\overline{z(t)} z'(t)\ dt.

(These formulas for perimeter and area are beautiful in their own right! Feel free to pause to admire them if you have not seen them before. đŸ™‚ )

Since we know that scaling the coordinates of the curve by k increases the arc length by k and its area enclosed by k^2 , we may assume that the perimeter is normalised to {P = 2\pi} and the curve is parametrised by arc length, so that {|z'(t)|=1}. Writing z(t) in terms of its Fourier series gives \displaystyle z(t) = \sum_{n \in \mathbb{Z}} c_n e^{int}.

By translating the curve by {c_0} if necessary (not changing the length or area enclosed) we may assume {c_0 = 0} and so

\displaystyle z(t) = \sum_{n \neq 0} c_n e^{int}.\quad ...(1)

Differentiating (1) term by term gives

\displaystyle z'(t) = \sum_{n\neq 0} c_n (in) e^{int} \quad ...(2)

(true by the smoothness assumption of z'(t)), and so

\begin{array}{lcl} A &=& \frac{1}{2i} \int_0^{2\pi}\overline{z(t)} z'(t)\ dt\\&=& \frac{1}{2i} \int_0^{2\pi} \sum_{m\neq 0} \overline{c_m} e^{-imt} \sum_{n \neq 0} c_n (in) e^{int} \ dt\\&=& \frac{1}{2}\sum_{m\neq 0}\sum_{n\neq 0}\overline{c_m}n c_n \int_0^{2\pi} e^{i(n-m)t}\ dt\\&=& \frac{1}{2}\sum_{m\neq 0}\sum_{n\neq 0}\overline{c_m}nc_n 2\pi \delta(n-m) \\&=& \pi\sum_{n\neq 0} n|c_n|^2. \quad ...(3)\end{array}

The interchange of summation and integration above is justified by the uniform convergence of the series. By Parseval’s relation on (2), we can write

\displaystyle \sum_{n \neq 0} n^2|c_n|^2 = \frac{1}{2\pi}\int_0^{2\pi} |z'(t)|^2\ dt = \frac{1}{2\pi}\int_0^{2\pi} 1\ dt = 1, \quad ...(4)

where the second equality follows from our assumption of z being parametrised by arc length. Using the elementary inequality {n \leq n^2} for {n \in \mathbb{Z}} from (3) and then (4) we have

\displaystyle A \leq \pi \sum_{n \neq 0} n^2 |c_n|^2 = \pi.

Since n is strictly less than n^2 unless n is 0 or 1, equality above holds if c_n = 0 for alln \neq 1, in which case z(t) has the Fourier series representation

\displaystyle z(t) = c_1 e^{it}.

Since |z'(t)| = 1 this means |c_1| = 1, so z(t) is the unit circle as expected.

By scaling by a factor of P/(2\pi), if z(t) has length P and encloses an area A, then A \leq \frac{P^2}{4 \pi}. This is the isoperimetric inequality.

References

[1] Alberto Candel, Notes on Fourier Series, available at http://www.csun.edu/~ac53971/courses/math650/fourier.pdf

[2] http://cornellmath.wordpress.com/2008/05/16/two-cute-proofs-of-the-isoperimetric-inequality/

[3] http://www.cut-the-knot.org/do_you_know/isoperimetric.shtml

[4] Andrejs Treibergs, Inequalities that Imply the Isoperimetric Inequality, available at http://www.math.utah.edu/~treiberg/isoperim/isop.pdf

[5] Alan Siegel, A Historical Review of the Isoperimetric Theorem in 2-D, and its place in Elementary Plane Geometry, available at http://www.cs.nyu.edu/faculty/siegel/SCIAM.pdf

Advertisements

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at WordPress.com.

%d bloggers like this: