Home > shape optimization > Proof of the isoperimetric inequality 2

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.

The proof I give here assumes again the existence of the optimal shape, a non-trivial fact proved here. Let’s make a few remarks about the optimal shape. Like in the proof presented here, it is reasonable to assume that the plane region enclosed by {L} denoted by {\Omega} is convex.

One of the basic applications of continuity to plane geometry to see that for any direction {\overrightarrow{d}} in the plane there exists a line parallel to this direction which splits {\Omega} in two region of equal areas. To do this, take a ‘sweep’ of the plane by a line parallel to {\overrightarrow{d}}, i.e. choose a coordinate system {xOy} in the plane such that {\Omega} is included in the halfplane {x>0}, and denote {f(x)=|\Omega \cap \{ y \in \Bbb{R}^2 : y<x \}|}, where {|\cdot|} is the Lebesgue measure. It can be proved that {f} is in fact Lipschitz continuous (note that {\Omega} is considered to be bounded). Since {f(0)=0} and {\lim_{x \rightarrow \infty}f(x)=|\Omega|}, the continuity of {f} implies that there exists {x_d >0} such that {f(x_d)=|\Omega|/2}, and the line {x=x_d} has the desired property.

Using the fact presented above and another continuity argument we can prove that there exists a line {\ell} which splits {\Omega} into two regions which have the same area and the same perimeter. For every direction which we can characterize by the angle {\theta \in (0,2\pi)} which this direction makes with the positive side of the axis {Ox} there exists a line {\ell_\theta} which splits {\Omega} in two regions of equal areas. For such a line {\ell_\theta} we denote by {P_+(\theta)} and {P_-(\theta)} the perimeter of {\Omega} on one side and the other of {\ell_\theta}. We choose the {+} and {-} signs such that the orientation is consistent with the case {\theta=0}. Then we can define {f:[0,2\pi]} by

\displaystyle f(\theta)= P_+(\theta)-P_-(\theta),

a function which is again continuous. We note that {f(\theta+\pi)=-f(\theta)} since after a rotation by {\pi} the line support of the direction is the same, the direction points the other way. Therefore we can find a {\theta \in [0,\pi]} such that {f(\theta)=0} and we are done.

Using this trick we can choose this line which splits {\Omega} into two parts with equal areas and equal perimeters and reduce the problem to proving that half the optimal shape is half a circle. For this denote {A,B} the intersection of {\ell} with {\Gamma} and pick one half of {\Omega}. Suppose that this half {\Omega'} is not a half-circle. Then we can find a point {C} on the boundary (not on {\ell}) such that the angle { \angle ACB} is less or more than {\pi/2}. The area of {ACB} is given by the formula

\displaystyle \text{Area} ABC=\frac{AC\cdot BC \cdot \sin \angle ACB}{2},

so if we ‘glue’ the portions of {\Omega' \setminus \Delta ABC} to the sides {AC,BC}, and consider a hinge in {C} such that these regions are mobile (see the figure). If we move the considered regions until {\angle C=\pi/2} then we have increased the area preserving the perimeter, and by the symmetry by {\ell} we get a contradiction to the fact that {\Omega} is optimal.



This contradiction shows that half of {\Omega} is half a circle, and therefore {\Omega} itself is a circle.


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