### Archive

Posts Tagged ‘isoperimetric’

## 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.

## Proof of the isoperimetric inequality 2

I will continue the series of proofs for the isoperimetric inequality in the two dimensional case, i.e. if a simple closed curve ${ \Gamma}$ (which we suppose for simplicity that it consists of piecewise ${ C^1}$ curves) of length ${ L}$ encloses an area ${ A}$ then ${ L^2 \geq 4\pi A}$ and the equality is attained only if ${ \Gamma}$ is the boundary of a circle.

Categories: shape optimization

## Proof of the Isoperimetric Inequality

The Isoperimetric inequality gives a bound for the area in terms of the perimeter of a set. It says that the greatest area that can be enclosed by a curve which has length $L$ is maximal when the curve is the boundary of a circle, or equivalently the minimum of the perimeter of a curve which encloses a set of area $A$ is attained again for the circle. In two dimension the inequality says that $L^2 \geq 4\pi A$ where $L$ is the perimeter and $A$ is the area (Lebesgue measure) of a plane region $\Omega$.

Categories: shape optimization

## Region which can sustain the Largest Sandpile

Among all plane regions $\Omega$ which are open and simply connected and without holes of given area $A$ the circle can support the largest sandpile.

Leavitt and Ungar – Circle supports the largest sandpile.

Categories: shape optimization

## Existence Result for the Isoperimetric Problems

The tricky part is how to define the perimeter of a Lebesgue measurable set with finite perimeter. This can be done considering the space of bounded variation functions, denoted $BV(\Bbb{R}^N)$. By definition we have for an open set $U \subset \Bbb{R}^N$ that $BV(U)=\left\{ f \in L^1(U) : \sup \left\{\int_U f {\rm div} \varphi dx | \varphi \in C_c^1(U;\Bbb{R}^N),\ |\varphi|\leq 1\right\} \right\}$.  Here we denoted by $C_c^1(U;\Bbb{R}^N)$ the space of continuously differentiable functions $f : U \to \Bbb{R}^N$ with compact support in $U$. Because of the density of the space $C_c^\infty(U,\Bbb{R}^N)$ of infinitely differentiable functions $f: U \to \Bbb{R}^N$ with compact support in $U$ in the space $C_c^1(U;\Bbb{R}^N)$, we could have replaced $C_c^1$ by $C_c^\infty$ in the above definition. You could take a look at this blog post for a detailed description of $BV(U)$ or at the Wikipedia page.
We say that a set $A$ of finite Lebesgue measure is a set of finite perimeter in $\Bbb{R}^N$ if its characteristic function $\chi_A$ belongs to $BV(\Bbb{R}^N)$. This means that the distributional gradient $\nabla \chi_A$ is a vector valued measure with finite total variation. The total variation $|\nabla \chi_A|$ is called the perimeter of $A$.
In the same way we can define the perimeter of a Lebesgue measurable set $A$ relative to an open set $D$. We say that $A\subset D$ is a set of finite perimeter relative to $D$ if the characteristic function $\chi_A$ belongs to the space $BV(D)$.