Hamilton's principle/Mathematics/Isoperimetric problems

Plain-language version: 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 Lagrange multipliers.

Statement

Fixed perimeter, varying area.

Theorem. Let $y\in 𝒜$ be stationary for

$J[y]={\int }_a^{b}L(x,y,y^{\prime })dx\mathrm{(}\mathrm{1}\mathrm{)}$

subject to

$K[y]={\int }_a^{b}M(x,y,y^{\prime })dx=\ell \mathrm{(}\mathrm{2}\mathrm{)}$

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

$L^{\ast }=L+\mu M\mathrm{(}\mathrm{3}\mathrm{)}$

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

Derivation

A two-parameter family for an isoperimetric problem.

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

$y_{{\epsilon }_1{\epsilon }_2}=y+{\epsilon }_1{\eta }_1+{\epsilon }_2{\eta }_2\mathrm{(}\mathrm{4}\mathrm{)}$

with ${\eta }_1,{\eta }_2\in 𝒱$. Two parameters give one to probe the functional and one to steer back onto the constraint. Define

$\Phi ({\epsilon }_1,{\epsilon }_2)=J[y_{{\epsilon }_1{\epsilon }_2}],\Psi ({\epsilon }_1,{\epsilon }_2)=K[y_{{\epsilon }_1{\epsilon }_2}]-\ell \mathrm{(}\mathrm{5}\mathrm{)}$

Both vanish appropriately at the origin: $\Psi (0,0)=0$ because $y$ satisfies the constraint.

The problem is now finite-dimensional: minimise $\Phi$ on the curve $\Psi =0$ in the $({\epsilon }_1,{\epsilon }_2)$ plane, at the origin. That is ordinary constrained calculus, and its multiplier rule gives a constant $\mu$ with

$\mathrm{\nabla }\Phi (0,0)+\mu \mathrm{\nabla }\Psi (0,0)=0\mathrm{(}\mathrm{6}\mathrm{)}$

$\mu$ 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 $\mathrm{\nabla }\Psi (0,0)\ne 0$, that is, some variation must actually change $K$ to first order. Failure of that is the abnormal case.

Writing the components of (6) out, with $\partial \Phi /\partial {\epsilon }_j{|}_0=\delta J[y;{\eta }_j]$ and similarly for $\Psi$,

$\delta J[y;{\eta }_j]+\mu \delta K[y;{\eta }_j]=0,j=1,2\mathrm{(}\mathrm{7}\mathrm{)}$

Since ${\eta }_2$ was arbitrary in $𝒱$ and $\mu$ does not depend on it, (7) holds for every $\eta \in 𝒱$. Expanding both variations and applying the fundamental lemma,

$\frac{\partial }{\partial y}(L+\mu M)-\frac{d}{dx}\frac{\partial }{\partial y^{\prime }}(L+\mu M)=0\mathrm{(}\mathrm{8}\mathrm{)}$

$◼$

Determining the multiplier

$\mu$ is an unknown of the problem, not an input. Equation (8) is second order, so its solution carries two constants; with $\mu$ that is three unknowns. The conditions available are $y(a)=y_a$, $y(b)=y_b$, 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 $\mu$. Solving for $\mu$ is usually the hardest step, and it is generally transcendental.

The abnormal case

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

Such a $y$ 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 $L$ and $M$ up to the factor $\mu$. Dividing through, the same extremals arise from extremising $M$ subject to fixed $J$. 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

the Beltrami identity derived the catenary while suppressing the length constraint. Restore it. Minimise potential energy

$J[y]={\int }_a^{b}y\sqrt{1+y{\text{'}}^2}dx\mathrm{(}\mathrm{9}\mathrm{)}$

subject to fixed length

$K[y]={\int }_a^b\sqrt{1+y{\text{'}}^2}dx=\ell \mathrm{(}\mathrm{10}\mathrm{)}$

By (3),

$L^{\ast }=(y+\mu )\sqrt{1+y{\text{'}}^2}\mathrm{(}\mathrm{11}\mathrm{)}$

$L^{\ast }$ contains no explicit $x$, so the Beltrami identity applies. It has the same form as the unconstrained case with $y$ replaced by $y+\mu$, so the same two-line reduction gives

$\frac{y+\mu }{\sqrt{1+y{\text{'}}^2}}=C\mathrm{(}\mathrm{12}\mathrm{)}$

and therefore

$y=-\mu +C\cosh \left(\frac{x-x_0}{C}\right)\mathrm{(}\mathrm{13}\mathrm{)}$

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

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

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

Several constraints

With $s$ isoperimetric constraints $K_{\gamma }[y]={\ell }_{\gamma }$, the family carries $s+1$ parameters and the argument gives $s$ constants:

$L^{\ast }=L+\sum _{\gamma =1}^s{\mu }_{\gamma }{M}_{\gamma }\mathrm{(}\mathrm{14}\mathrm{)}$

The non-degeneracy hypothesis becomes the requirement that the $\delta K_{\gamma }$ 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:

$L^{\ast }=L+\sum _{\beta }{\lambda }_{\beta }(x)g_{\beta }+\sum _{\gamma }{\mu }_{\gamma }{M}_{\gamma }\mathrm{(}\mathrm{15}\mathrm{)}$

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

Next

the second variation leaves constraints behind and asks the question the machinery track deferred: whether an extremal is a minimum at all.