Mathematical Methods for Physics and Engineering : A Comprehensive Guide

CALCULUS OF VARIATIONS

−a

y

O a

x

Figure 22.7 A uniform rope with fixed end-points suspended under gravity.

which, together with the original constraintJ= constant, will yield the required

solutiony(x).

This method is easily generalised to cases with more than one constraint by the

introduction of more Lagrange multipliers. If we wish to find the stationary values

of an integralIsubject to the multiple constraints that the values of the integrals

Jibe held constant fori=1, 2 ,...,n, then we simply find the unconstrained

stationary values of the new integral

K=I+

∑n

1

λiJi.

Find the shape assumed by a uniform rope when suspended by its ends from two points

at equal heights.

We will solve this problem usingx(see figure 22.7) as the independent variable. Let
theropeoflength2Lbe suspended between the pointsx=±a,y=0(L>a)and
have uniform linear densityρ. We then need to find the stationary value of the rope’s
gravitational potential energy,

I=−ρg

∫

yds=−ρg

∫a

−a

y(1 +y′^2 )^1 /^2 dx,

with respect to small changes in the form of the rope but subject to the constraint that
the total length of the rope remains constant, i.e.

J=

∫

ds=

∫a

−a

(1 +y′^2 )^1 /^2 dx=2L.

We thus define a new integral (omitting the factor−1fromIfor brevity)

K=I+λJ=

∫a

−a

(ρgy+λ)(1 +y′^2 )^1 /^2 dx

and find its stationary values. Since the integrand does not contain the independent
variablexexplicitly, we can use (22.8) to find the first integral:

(ρgy+λ)

(

1+y′^2

) 1 / 2

−(ρgy+λ)

(

1+y′^2

)− 1 / 2

y′^2 =k,