Isoperimetric Inequality

Minimum length of the curve bounding an area A? Maximum area bound by a closed curve of length L? Optimal case: Circle. L^2 \ge 4\pi A

Steiner Proof: Consider a curve with given perimeter L. 1. If the region bound by the curve is not convex – there is a chord joining two point on the curve which lies outside the area bound. Reflecting the portion of curve about that chord we preserve the perimeter but increase the area of the bound region. 2. Split the area into two parts by dividing the length into two equal parts- that is take two points which are at a distance L/2 along the curve and join them. These parts need to have the same area- otherwise just reflect the larger part and we increase the area. 3. Each of these parts has to be a semicircle. If it’s not there is a triangle PAB with three which is not right angled at P. (P is on the curve, AB is the diameter-which divided the curve into two parts). By just deforming the angle to 90, we increase the area of the triangle and all other lengths, areas are preserved- which ultimately increases the area.

Stokes theorem proof: Stokes theorem helps us to relate integrated local quantities to integrals of quantities on the boundary- so we should expect Stokes theorem to see the relation between the area and perimeter of a curve.

We have \displaystyle A=-\int_{C} y d x=-\int_{a}^{b} y \frac{d x}{d t} d t

\displaystyle t=(2 \pi / L) s, where s denotes arclength parameter.

\displaystyle \begin{aligned} L^{2}-4 \pi A &=2 \pi \int_{0}^{2 \pi}\left[\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}+2 y \frac{d x}{d t}\right] d t \\ &=2 \pi \int_{0}^{2 \pi}\left(\frac{d x}{d t}+y\right)^{2} d t+2 \pi \int_{0}^{2 \pi}\left[\left(\frac{d y}{d t}\right)^{2}-y^{2}\right] d t \end{aligned}

y(t) is a periodic function with no constant term- use the Fourier expansion to see that the above expression is positive!

Another idea: We can directly compare the area expressions for our curve and a circle with same diameter- Applying Cauchy Schwarz- we can see that circle has to be the optimal case.


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