Monday Math 1: The Isoperimetric Problem

Isoperimetric Problem

The isoperimetric problem is the problem of finding the closed plane curve with a given perimeter that encloses the greatest area. Below, I will use the calculus of variations in a proof of the long-known solution, the circle.

We let our enclosed region be D, and the enclosing curve ∂D, with orientation on ∂D chosen so that D is to it's left (counterclockwise). We thus want to minimize $A=\iint_D\,dxdy$ subject to the constraint $p=\oint_{\partial D}\,ds$ . Note that these are quite different kinds of integrals, and cannot be combined directly via the multiplier method of including constraints.

However, using Green's theorem, which says that over a plane region D with (counterclockwise oriented) boundary ∂D we have:
$\oint_{\partial D}f(x,y)dx+g(x,y)dy=\iint_D\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,dxdy$

By choosing $f(x,y)=-y,\,g(x,y)=x$ , and using the fact that the area of D is $A=\iint_D\,dxdy$ , we obtain $A=\frac{1}{2}\oint_{\partial D}xdy-ydx$

So now we can combine the minimized quantity and the constraint using a multiplier. As both are integral constraints, the multiplier is a constant λ, so parametrizing our curve ∂D as (x(t),y(t)), so
$ds=\sqrt{dx^2+dy^2}=\sqrt{{x'}^2+{y'}^2}$ and we want to find the extrema of

$S=\oint_{\partial D}\frac{1}{2}(xy'-yx')+\lambda\sqrt{{x'}^2+{y'}^2}\,dt$

The integrand is $L=\frac{1}{2}(xy'-yx')+\lambda\sqrt{{x'}^2+{y'}^2}$ , so we have the Euler-Lagrange equation for x:
$\frac{\partial L}{\partial x}-\frac{d}{dt}\left(\frac{\partial L}{\partial x'}\right)=0$
$\frac{1}{2}y'-\frac{d}{dt}\left(-\frac{y}{2}+\frac{\lambda x'}{\sqrt{{x'}^2+{y'}^2}}\right)=0$
Now, $y'=\frac{dy}{dt}$ , so $\frac{1}{2}y'=\frac{d}{dt}\left(\frac{y}{2}\right)$ , and
$\frac{d}{dt}\left(y-\frac{\lambda x'}{\sqrt{{x'}^2+{y'}^2}}\right)=0$
$y-\frac{\lambda x'}{\sqrt{{x'}^2+{y'}^2}}=C_1$
Similarly for y:
$\frac{\partial L}{\partial y}-\frac{d}{dt}\left(\frac{\partial L}{\partial y'}\right)=0$
$-\frac{1}{2}x'-\frac{d}{dt}\left(\frac{x}{2}+\frac{\lambda y'}{\sqrt{{x'}^2+{y'}^2}}\right)=0$
$\frac{d}{dt}\left(-x-\frac{\lambda y'}{\sqrt{{x'}^2+{y'}^2}}\right)=0$
$x+\frac{\lambda y'}{\sqrt{{x'}^2+{y'}^2}}=C_2$

So we have $x-C_2=-\frac{\lambda y'}{\sqrt{{x'}^2+{y'}^2}}$ and $y-C_1=\frac{\lambda x'}{\sqrt{{x'}^2+{y'}^2}}$
Squaring and summing these, we obtain:
$(x-C_2)^2+(y-C_1)^2=\frac{\lambda^2{y'}^2}{{x'}^2+{y'}^2}+\frac{\lambda^2{x'}^2}{{x'}^2+{y'}^2}$
$(x-C_2)^2+(y-C_1)^2=\lambda^2$

This is the equation of a circle with radius λ and and center (C2,C1). From our original constraint $p=\oint_{\partial D}\,ds$ , we obtain λ=p/2π.

Like this:

Be the first to like this post.

This entry was posted on December 31, 2007 at 3:33 pm and is filed under Uncategorized. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Monday Math 1: The Isoperimetric Problem”

Physics Friday 65 « Twisted One 151’s Weblog Says:
March 27, 2009 at 12:10 am | Reply
[...] that , and to this integral over our closed curve (c), we apply Green’s Theorem (see also here), which says , where D is the plane region bound by the simple closed curve ∂D. Here, [...]
Physics Friday 124 « Twisted One 151's Weblog Says:
June 18, 2010 at 12:26 am | Reply
[...] our integral for magnetic moment becomes , and via Green’s theorem (reversing our use of it here), we see , where n is the normal to the plane of the loop (with direction via the right-hand rule), [...]

Twisted One 151's Weblog