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	{{Plebeian\|Hamilton's principle/Explainers/Fixed length problems}}		{{Plebeian\|Hamilton's principle/Explainers/Fixed length problems}}

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{{Plebeian|Hamilton's principle/Explainers/Fixed length problems}}

An '''isoperimetric problem''' asks for the extremum of one integral while a
second integral is held at a prescribed value. The name comes from the oldest
instance — enclose the greatest area with a curve of given perimeter — and the
class contains the hanging chain, the shape of a soap film with fixed volume, and
the ground state of a quantum system with fixed normalisation.

The constraint here is a single number for the whole path, not a relation at each
point. That difference, and not the algebra, is what separates this page from
[[Hamilton's principle/Mathematics/Lagrange multipliers]].

== Statement ==

[[File:OPL-HP-F-20.png|thumb|right|400px|Fixed perimeter, varying area.]]

'''Theorem.''' Let <math>y \in \mathcal{A}</math> be stationary for

:<math>J[y] = \int_a^b L(x, y, y')\, dx
\qquad\qquad\mathrm{(1)}</math>

subject to

:<math>K[y] = \int_a^b M(x, y, y')\, dx = \ell
\qquad\qquad\mathrm{(2)}</math>

for a prescribed constant <math>\ell</math>. If <math>y</math> is not itself an
extremal of <math>K</math> alone, then there is a '''constant''' <math>\mu</math>
such that <math>y</math> satisfies the Euler-Lagrange equation of

:<math>L^* = L + \mu M
\qquad\qquad\mathrm{(3)}</math>

The excluded case is called '''abnormal''' and is discussed below. It is a real
exclusion, not a technicality.

== Derivation ==

[[File:OPL-HP-F-35.png|thumb|right|400px|A two-parameter family for an isoperimetric problem.]]

A single one-parameter family will not do. Varying along
<math>y + \varepsilon\eta</math> generally changes <math>K</math>, so the
competitor leaves the constraint set and the comparison is meaningless. The fix
is a '''two-parameter''' family:

:<math>y_{\varepsilon_1 \varepsilon_2}
= y + \varepsilon_1\eta_1 + \varepsilon_2\eta_2
\qquad\qquad\mathrm{(4)}</math>

with <math>\eta_1, \eta_2 \in \mathcal{V}</math>. Two parameters give one to
probe the functional and one to steer back onto the constraint. Define

:<math>\Phi(\varepsilon_1,\varepsilon_2) = J[y_{\varepsilon_1\varepsilon_2}],
\qquad
\Psi(\varepsilon_1,\varepsilon_2) = K[y_{\varepsilon_1\varepsilon_2}] - \ell
\qquad\qquad\mathrm{(5)}</math>

Both vanish appropriately at the origin: <math>\Psi(0,0) = 0</math> because
<math>y</math> satisfies the constraint.

The problem is now finite-dimensional: minimise <math>\Phi</math> on the curve
<math>\Psi = 0</math> in the <math>(\varepsilon_1, \varepsilon_2)</math> plane,
at the origin. That is ordinary constrained calculus, and its multiplier rule
gives a constant <math>\mu</math> with

:<math>\nabla\Phi(0,0) + \mu\,\nabla\Psi(0,0) = 0
\qquad\qquad\mathrm{(6)}</math>

<math>\mu</math> is a single number because (6) is a statement at one point of a
plane, not at every point of an interval. '''This is the entire reason the
multiplier is constant here and a function on the previous page.'''

The rule requires <math>\nabla\Psi(0,0) \neq 0</math>, that is, some variation
must actually change <math>K</math> to first order. Failure of that is the
abnormal case.

Writing the components of (6) out, with
<math>\partial\Phi/\partial\varepsilon_j |_0 = \delta J[y;\eta_j]</math> and
similarly for <math>\Psi</math>,

:<math>\delta J[y;\eta_j] + \mu\,\delta K[y;\eta_j] = 0,
\qquad j = 1, 2
\qquad\qquad\mathrm{(7)}</math>

Since <math>\eta_2</math> was arbitrary in <math>\mathcal{V}</math> and
<math>\mu</math> does not depend on it, (7) holds for every
<math>\eta \in \mathcal{V}</math>. Expanding both variations and applying the
fundamental lemma,

:<math>\frac{\partial}{\partial y}(L + \mu M)
- \frac{d}{dx}\frac{\partial}{\partial y'}(L + \mu M) = 0
\qquad\qquad\mathrm{(8)}</math>

<math>\blacksquare</math>

== Determining the multiplier ==

<math>\mu</math> is an unknown of the problem, not an input. Equation (8) is
second order, so its solution carries two constants; with <math>\mu</math> that
is three unknowns. The conditions available are <math>y(a) = y_a</math>,
<math>y(b) = y_b</math>, and the constraint (2) itself. Three conditions, three
unknowns.

The constraint is therefore used twice: once to justify the method, and once at
the end to pin <math>\mu</math>. Solving for <math>\mu</math> is usually the
hardest step, and it is generally transcendental.

== The abnormal case ==

If <math>y</math> is an extremal of <math>K</math> alone, then
<math>\delta K[y;\eta] = 0</math> for every <math>\eta</math>,
<math>\nabla\Psi(0,0) = 0</math>, and (6) collapses. Geometrically, the
constraint set has a singular point at <math>y</math> and there is no tangent
direction to move along.

Such a <math>y</math> can still be a legitimate solution — it may be the ''only''
path meeting the constraint — but it is not found by (8), and a treatment that
omits the hypothesis will silently miss or misdescribe it. The exclusion is
standard and is stated here because it is often dropped.

== Symmetry of the two integrals ==

Equation (8) is symmetric in <math>L</math> and <math>M</math> up to the factor
<math>\mu</math>. Dividing through, the same extremals arise from extremising
<math>M</math> subject to fixed <math>J</math>. This is the '''reciprocity''' of
isoperimetric problems: maximising area at fixed perimeter and minimising
perimeter at fixed area have the same solutions. The classical result that the
circle solves both is one theorem, read twice.

== Worked example: the catenary, done properly ==

[[Hamilton's principle/Mathematics/The Beltrami identity]] derived the catenary
while suppressing the length constraint. Restore it. Minimise potential energy

:<math>J[y] = \int_a^b y\sqrt{1 + y'^2}\; dx
\qquad\qquad\mathrm{(9)}</math>

subject to fixed length

:<math>K[y] = \int_a^b \sqrt{1 + y'^2}\; dx = \ell
\qquad\qquad\mathrm{(10)}</math>

By (3),

:<math>L^* = (y + \mu)\sqrt{1 + y'^2}
\qquad\qquad\mathrm{(11)}</math>

<math>L^*</math> contains no explicit <math>x</math>, so the Beltrami identity
applies. It has the same form as the unconstrained case with <math>y</math>
replaced by <math>y + \mu</math>, so the same two-line reduction gives

:<math>\frac{y + \mu}{\sqrt{1 + y'^2}} = C
\qquad\qquad\mathrm{(12)}</math>

and therefore

:<math>y = -\mu + C\cosh\!\left( \frac{x - x_0}{C} \right)
\qquad\qquad\mathrm{(13)}</math>

The curve is the same catenary, shifted vertically by <math>\mu</math>. Three
constants <math>C</math>, <math>x_0</math>, <math>\mu</math> are fixed by the two
endpoints and by (10).

The multiplier here has a reading: <math>-\mu</math> is the height of the
directrix, the level from which the chain's tension is measured. The horizontal
tension component is proportional to <math>C</math>. Neither number was available
from the suppressed-constraint treatment.

Whether the chain is longer than the straight-line distance matters. If
<math>\ell</math> equals that distance the only admissible path is the straight
segment, which is an extremal of <math>K</math> alone — the abnormal case,
exactly as the exclusion predicts.

== Several constraints ==

With <math>s</math> isoperimetric constraints
<math>K_\gamma[y] = \ell_\gamma</math>, the family carries <math>s+1</math>
parameters and the argument gives <math>s</math> constants:

:<math>L^* = L + \sum_{\gamma=1}^{s} \mu_\gamma M_\gamma
\qquad\qquad\mathrm{(14)}</math>

The non-degeneracy hypothesis becomes the requirement that the
<math>\delta K_\gamma</math> be linearly independent as functionals.

== Mixed problems ==

Nothing prevents a problem from carrying both pointwise and isoperimetric
constraints. The augmented integrand then contains multiplier ''functions'' for
the first and multiplier ''constants'' for the second:

:<math>L^* = L
+ \sum_\beta \lambda_\beta(x)\, g_\beta
+ \sum_\gamma \mu_\gamma\, M_\gamma
\qquad\qquad\mathrm{(15)}</math>

Getting the function-versus-constant distinction wrong in (15) is the commonest
error in the subject, and the test remains the one stated on
[[Hamilton's principle/Mathematics/Lagrange multipliers]]: one equation per
point, or one equation for the path.

== Next ==

[[Hamilton's principle/Mathematics/The second variation]] leaves constraints
behind and asks the question the machinery track deferred: whether an extremal
is a minimum at all.

== References ==

<references />

← Older revision		Revision as of 04:57, 16 September 2026
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	The constraint here is a single number for the whole path, not a relation at each		The constraint here is a single number for the whole path, not a relation at each
	point. That difference, and not the algebra, is what separates this page from		point. That difference, and not the algebra, is what separates this page from
	[[Hamilton's principle/Mathematics/Lagrange multipliers]].		[[Hamilton's principle/Mathematics/Lagrange multipliers\|Lagrange multipliers]].

	== Statement ==		== Statement ==
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	== Worked example: the catenary, done properly ==		== Worked example: the catenary, done properly ==

	[[Hamilton's principle/Mathematics/The Beltrami identity]] derived the catenary		[[Hamilton's principle/Mathematics/The Beltrami identity\|the Beltrami identity]] derived the catenary
	while suppressing the length constraint. Restore it. Minimise potential energy		while suppressing the length constraint. Restore it. Minimise potential energy

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	Getting the function-versus-constant distinction wrong in (15) is the commonest		Getting the function-versus-constant distinction wrong in (15) is the commonest
	error in the subject, and the test remains the one stated on		error in the subject, and the test remains the one stated on
	[[Hamilton's principle/Mathematics/Lagrange multipliers]]: one equation per		[[Hamilton's principle/Mathematics/Lagrange multipliers\|Lagrange multipliers]]: one equation per
	point, or one equation for the path.		point, or one equation for the path.

	== Next ==		== Next ==

	[[Hamilton's principle/Mathematics/The second variation]] leaves constraints		[[Hamilton's principle/Mathematics/The second variation\|the second variation]] leaves constraints
	behind and asks the question the machinery track deferred: whether an extremal		behind and asks the question the machinery track deferred: whether an extremal
	is a minimum at all.		is a minimum at all.

Hamilton's principle/Mathematics/Isoperimetric problems - Revision history

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