Mathematical Methods for Physics and Engineering : A Comprehensive Guide

CALCULUS OF VARIATIONS

A

y

x=x 0

B

x

Figure 22.6 A frictionless wire along which a small bead slides. We seek the

shape of the wire that allows the bead to travel from the originOto the line

x=x 0 in the least possible time.

Now, from (22.17) the conditionδI= 0 requires, besides the EL equation, that

atx=b, the other two contributions cancel, i.e.

F∆x+

∂F

∂y′

η=0. (22.19)

Eliminating ∆xandηbetween (22.18) and (22.19) leads to the condition that at

the end-point
(
F−y′

∂F

∂y′

)

∂h

∂y

−

∂F

∂y′

∂h

∂x

=0. (22.20)

In the special case where the end-point is free to lie anywhere on the vertical line

x=b, we have∂h/∂x= 1 and∂h/∂y= 0. Substituting these values into (22.20),

we recover the end-point condition given in (22.16).

A frictionless wire in a vertical plane connects two pointsAandB,Abeing higher thanB.

Let the position ofAbe fixed at the origin of anxy-coordinate system, but allowBto lie

anywhere on the vertical linex=x 0 (see figure 22.6). Find the shape of the wire such that

abeadplacedonitatAwill slide under gravity toBin the shortest possible time.

This is a variant of the famous brachistochrone (shortest time) problem, which is often
used to illustrate the calculus of variations. Conservation of energy tells us that the particle
speed is given by

v=

ds

dt

=

√

2 gy,

wheresis the path length along the wire andgis the acceleration due to gravity. Since
the element of path length isds=(1+y′^2 )^1 /^2 dx, the total time taken to travel to the line
x=x 0 is given by

t=

∫x=x 0

x=0

ds

v

=

1

√

2 g

∫x 0

0

√

1+y′^2

y

dx.

Because the integrand does not containxexplicitly, we can use (22.8) with the specific
formF=

√

1+y′^2 /

√

yto find a first integral; on simplification this yields

[

y(1 +y′^2 )

] 1 / 2

=k,