Constrained variation

22.4 CONSTRAINED VARIATION
where k is a constant. Letting a = k 2 and solving for y we find
y =
a−y
,
y
dy
=
dx
which on substituting y = a sin2 θ integrates to give
a
x = (2θ − sin 2θ) + c.
2
Thus the parametric equations of the curve are given by
x = b(φ − sin φ) + c,
y = b(1 − cos φ),
where b = a/2 and φ = 2θ; they define a cycloid, the curve traced out by a point on
the rim of a wheel of radius b rolling along the x-axis. We must now use the end-point
conditions to determine the constants b and c. Since the curve passes through the origin,
we see immediately that c = 0. Now since y(x0 ) is arbitrary, i.e. the upper end-point can
lie anywhere on the curve x = x0 , the condition (22.20) reduces to (22.16), so that we also
require
∂F y
=
= 0,
2 ∂y x=x0
y(1 + y ) x=x
0
which implies that y = 0 at x = x0 . In words, the tangent to the cycloid at B must be
parallel to the x-axis; this requires πb = x0 . 22.4 Constrained variation
Just as the problem of finding thestationary values of a function f(x, y) subject to
the constraint g(x, y) = constant is solved by means of Lagrange’s undetermined
multipliers (see chapter 5), so the corresponding problem in the calculus of
variations is solved by an analogous method.
Suppose that we wish to find the stationary values of
b
F(y, y , x) dx,
I=
a
subject to the constraint that the value of
b
G(y, y , x) dx
J=
a
is held constant. Following the method of Lagrange undetermined multipliers let
us define a new functional
b
(F + λG) dx,
K = I + λJ =
a
and find its unconstrained stationary values. Repeating the analysis of section 22.1
we find that we require
∂G
d ∂F
d ∂G
∂F
−
−
+
λ
= 0,
∂y
dx ∂y ∂y
dx ∂y 785
CALCULUS OF VARIATIONS
y
−a
a
O
x
Figure 22.7 A uniform rope with fixed end-points suspended under gravity.
which, together with the original constraint J = constant, will yield the required
solution y(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 integral I subject to the multiple constraints that the values of the integrals
Ji be held constant for i = 1, 2, . . . , n, then we simply find the unconstrained
stationary values of the new integral
K=I+
n
λi Ji .
1
Find the shape assumed by a uniform rope when suspended by its ends from two points
at equal heights.
We will solve this problem using x (see figure 22.7) as the independent variable. Let
the rope of length 2L be suspended between the points x = ±a, y = 0 (L > a) and
have uniform linear density ρ. We then need to find the stationary value of the rope’s
gravitational potential energy,
a
2
I = −ρg y ds = −ρg
y(1 + y )1/2 dx,
−a
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.
a
2
(1 + y )1/2 dx = 2L.
J=
ds =
−a
We thus define a new integral (omitting the factor −1 from I for brevity)
a
2
(ρgy + λ)(1 + y )1/2 dx
K = I + λJ =
−a
and find its stationary values. Since the integrand does not contain the independent
variable x explicitly, we can use (22.8) to find the first integral:
1/2
−1/2
2
2
2
− (ρgy + λ) 1 + y y = k,
(ρgy + λ) 1 + y 786