classical isoperimetric problem

The points $a$ and $b$ on the $x$-axis have to be by an arc with a given length (http://planetmath.org/ArcLength) $l$ such that the area between the $x$-axis and the arc is as great as possible.

Denote the equation of the searched arc by  $y=y(x)$.  The task, which belongs to the isoperimetric problems (http://planetmath.org/IsoperimetricProblem), can be formulated as

$\text{to maximise}\mathit{\hspace{1em}}{\displaystyle {\int }_a^b}y𝑑x$ (1)

under the constraint condition

${\displaystyle {\int }_a^b}\sqrt{1+y^{\prime \mathrm{\hspace{0.17em}2}}}𝑑x=l.$ (2)

We have the integrands

$$f(x,y,y^{\prime })\equiv y,g(x,y,y^{\prime })\equiv \sqrt{1+y^{\prime \mathrm{\hspace{0.17em}2}}}.$$

The variation problem for the functional in (1) may be considered as a free variation problem (without conditions) for the functional ${\int }_a^b(f-\lambda g)𝑑x$ where $\lambda$ is a Lagrange multiplier.  For this end we need the Euler–Lagrange differential equation (http://planetmath.org/EulerLagrangeDifferentialEquation)

${\displaystyle \frac{\partial }{\partial y}}(f-\lambda g)-{\displaystyle \frac{d}{dx}}{\displaystyle \frac{\partial }{\partial y^{\prime }}}(f-\lambda g)=\mathrm{\hspace{0.33em}0}.$ (3)

Since the expression $f-\lambda g$ does not depend explicitly on $x$, the differential equation (3) has, by the Beltrami identity, a first integral of the form

$$(f-\lambda g)-y^{\prime }\cdot (f_{y^{\prime }}^{\prime }-\lambda g_{y^{\prime }}^{\prime })=C_2,$$

which reads simply

$$y-\frac{\lambda }{\sqrt{1+y^{\prime \mathrm{\hspace{0.17em}2}}}}=C_2.$$

This differential equation may be written

$$y^{\prime }\equiv \frac{dy}{dx}=\frac{\sqrt{{\lambda }^2-{(y-C_2)}^2}}{y-C_2},$$

where one can separate the variables (http://planetmath.org/SeparationOfVariables) and integrate, obtaining the equation

$${(x-C_1)}^2+{(y-C_2)}^2={\lambda }^2$$

of a circle.  Here, the parametres $C_1,C_2,\lambda$ may be determined from the conditions

$$y(a)=y(b)=\mathrm{\hspace{0.33em}0},\text{arc length}=l.$$

Thus the extremal of this variational problem is a circular arc (http://planetmath.org/CircularSegment) connecting the given points.

Note that in every point of the arc, the angle of view of the line segment between the given points is constant.

Title	classical isoperimetric problem
Canonical name	ClassicalIsoperimetricProblem
Date of creation	2013-03-22 19:10:26
Last modified on	2013-03-22 19:10:26
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	22
Author	pahio (2872)
Entry type	Example
Classification	msc 47A60
Classification	msc 49K22
Classification	msc 49K05
Related topic	LagrangeMultiplierMethod
Related topic	CircularSegment
Related topic	AngleOfViewOfALineSegment