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.