Home > Curves & Surfaces, Differential Geometry, shape optimization > Proof of the Isoperimetric Inequality 3

Proof of the Isoperimetric Inequality 3

I will present here a third proof for the planar Isoperimetric Inequality, using some simple notions of differential curves. For this suppose that the simple closed plane curve {C} has length {L} and encloses area {A}. Then

\displaystyle L^2 \geq 4 \pi A

and equality holds if and only if {C} is a circle.

Choose parallel lines {\ell_1} and {\ell_2} tangent to {C} and enclosing {C}, as seen in the picture. We draw a circle {C_1} of radius {R} tangent to {\ell_1,\ell_2} and put the origin at its center, with the {y}-axis parallel to {\ell_i}. We parametrize {C} by arclength {\alpha(s)=(x(s),y(s)),\ s \in [0,L]} taking {\alpha(0) \in \ell_1} and {\alpha(s_0) \in \ell_2}. We consider {\alpha' :[0,L] \rightarrow \Bbb{R}^2} defined by

\displaystyle \alpha_1(s)=(x_1(s),y_1(s))=\begin{cases} (x(s),-\sqrt{R^2-x^2(s)} & 0\leq s \leq s_0 \\ (x(s),\sqrt{R^2-x^2(s)} & s_0 \leq s \leq L \end{cases}

Note that the image of {\alpha_1} is the circle {C_1}, but {\alpha_1} does not need to be a parametrization of {C_1} since it may cover some parts multiple times. Also note that {x(s)=x_1(s),\ 0\leq s \leq L}.

We use Green’s formula to obtain that the areas {A,A_1} of {C} and {C_1}, respectively are

\displaystyle A=\int_0^L x(s)y'(s)ds

\displaystyle A_1=\pi R^2=-\int_0^L y_1(s)x_1'(s)ds=-\int_0^L y_1(s)x'(s)ds.

We add these equations and apply the Cauchy-Schwarz inequality:

\displaystyle A+\pi R^2=\int_0^L (x(s)y'(s)-y_1(s)x'(s))ds =\int_0^L (x(s),y_1(s))\cdot (y'(s),-x'(s))ds \leq

\displaystyle \leq \int_0^L \|(x(s),y_1(s)\| \|=RL,

because {(x(s),y_1(s))} is always on the circle {C_1} of radius {R} centered at the origin, and {\|(x'(s),y'(s)\|=1} since {\alpha} is parametrized by arclength.

Therefore by applying the AM-GM inequality we have

\displaystyle RL\geq A+\pi R^2 \geq 2 \sqrt{\pi A R^2}

which is equivalent to

\displaystyle L^2 \geq 4 \pi A.

Suppose now that the equality holds. Then we have {A=\pi R^2} and {L=2 \pi R}, so {R} does not depend on the orientation of the parallel lines {\ell_1,\ell_2}. Furthermore, the equality holds in the Cauchy-Schwarz inequality and we have {(x(s),y_1(s))=R(y'(s),-x'(s))} (because one vector has length {R} and the other one has length {1}). This implies {x(s)=Ry'(s)}.

Now consider the case when the direction of {\ell_1,\ell_2} is changed by an angle equal to {\pi/2}. Then {x(s)} and {y(s)} change places in the parametrization and in the new coordinates. We could pick the position of the circle {C_1} such that {C_1} is tangent to all four lines {\ell_1,\ell_2} and their rotates with {\pi/2}. Therefore the change of coordinates is just swapping {x} and {y}. Reasoning the same way we get {y(s)=Rx'(s)}, and therefore {x^2(s)+y^2(s)=R^2} and {C} is a circle.

  1. No comments yet.
  1. No trackbacks yet.

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 )

Google+ photo

You are commenting using your Google+ 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 )


Connecting to %s

%d bloggers like this: