Calculus of Variations: Constrained Variations and Isoperimetric Problems

The Euler–Lagrange equation developed in CV04 assumes that the admissible function may be perturbed in every sufficiently small endpoint-preserving direction. Many important variational problems impose an additional global condition. A curve may be required to have a prescribed length, a density may have a prescribed total mass, a wavefunction may be normalized, or a trajectory may be required to satisfy a fixed integral budget. Such problems are examples of constrained variational problems.

The classical prototype is an isoperimetric problem: extremize one functional while another integral functional is held fixed. The central tool is the variational analogue of the finite-dimensional Lagrange-multiplier method. Under a normality condition, a constrained extremal of

∫ b J [y] = F(x,y, y′) dx a

(1)

subject to

∫ b K [y] = G (x,y,y ′)dx = k0 a

(2)

is an unconstrained stationary function of an augmented functional with integrand

H = F + λG,

(3)

where $λ$ is a constant multiplier. Classical treatments of this theorem and of isoperimetric problems may be found in [1, 2, 3, 4].

1 Why an integral constraint changes the variation

Suppose the endpoint values are fixed and the admissible class also satisfies

K [y] = k0.

(4)

For the ordinary perturbation

y𝜖 = y + 𝜖η,

(5)

we have

K [y𝜖] = K [y] + 𝜖 δK [y;η ] + o(𝜖).

(6)

Therefore an arbitrary endpoint-preserving $η$ generally leaves the constraint surface. To remain feasible to first order, it would have to satisfy

δK [y;η] = 0.

(7)

This means the variations available to the constrained problem form only the tangent directions to the constraint set. We can no longer assert that the first variation of $J$ vanishes for every endpoint-preserving test function.

Figure. A constrained extremum is taken over the level set $K [y] = k0$ , not over the whole function space. At the extremum, admissible first-order variations are tangent to the constraint set. The multiplier theorem says that the first variations of $J$ and $K$ are linearly dependent there.

2 Finite-dimensional analogy

For a function $f(x)$ constrained by

g(x ) = c,

(8)

an ordinary constrained extremum satisfies

∇f + λ ∇g = 0

(9)

provided $∇g ⁄= 0$ . The geometric reason is that $∇g$ is normal to the constraint surface while the directional derivative of $f$ vanishes in every tangent direction. Hence $∇f$ must also be normal to the surface and therefore parallel to $∇g$ .

In the calculus of variations, the gradients are replaced by first-variation functionals. The corresponding statement is

|----------------------------------------------------------------------| |δJ[y∗;η] + λδK [y∗;η] = 0 for every admissible endpoint -preserving η.| -----------------------------------------------------------------------

(10)

The important point is that the multiplier restores the freedom to test against arbitrary endpoint-preserving directions.

3 The classical one-constraint multiplier theorem

Theorem. Let

∫ b ∫ b J [y] = F (x, y,y′)dx, K [y ] = G(x, y,y′) dx, a a

(11)

with $F$ and $G$ sufficiently smooth. Let $y∗ ∈ C2([a,b])$ satisfy fixed endpoint conditions and the constraint $K [y∗] = k0$ . Suppose $y∗$ is a local constrained extremum of $J$ . Assume the constraint is normal at $y ∗$ : there exists at least one endpoint-preserving variation $ζ$ such that

δK [y∗;ζ] ⁄= 0.

(12)

Then there exists a constant $λ$ such that

δJ [y∗;η] + λδK [y∗;η] = 0

(13)

for every endpoint-preserving variation $η$ .

3.1 Proof by a two-parameter variation

Take any endpoint-preserving test function $η$ , and use the special direction $ζ$ from the normality assumption. Form

y𝜖,δ = y ∗ + 𝜖η + δζ.

(14)

Define two ordinary functions of the two scalar parameters:

Φ(𝜖,δ) = J [y𝜖,δ],

(15)

and

Ψ (𝜖,δ ) = K [y ] − k . 𝜖,δ 0

(16)

At the origin,

Ψ (0, 0) = 0.

(17)

Moreover,

Ψδ(0,0) = δK [y∗;ζ] ⁄= 0.

(18)

The implicit-function theorem therefore gives, locally, a function $δ = h (𝜖)$ such that

Ψ(𝜖,h(𝜖)) = 0.

(19)

Thus

y𝜖,h(𝜖)

(20)

is a one-parameter family that stays on the constraint surface.

Figure. The two-parameter family supplies one arbitrary direction $η$ and one correction direction $ζ$ . The constraint equation determines a small correction $δ = h(𝜖)$ , producing a feasible curve through parameter space.

Because $y∗$ is a constrained extremum,

| d | --Φ (𝜖,h(𝜖))|| = 0. d𝜖 𝜖=0

(21)

By the ordinary chain rule,

Φ 𝜖(0,0 ) + Φ δ(0,0)h′(0) = 0.

(22)

Differentiating the constraint relation gives

′ Ψ 𝜖(0,0 ) + Ψ δ(0,0)h (0 ) = 0,

(23)

Ψ (0,0) h′(0) = − --𝜖-----. Ψ δ(0,0)

(24)

Substitution yields

Φδ(0,0)- Φ 𝜖(0,0) − Ψ δ(0,0) Ψ𝜖(0,0) = 0.

(25)

Recognize the parameter derivatives as first variations:

Φ 𝜖(0, 0) = δJ[y∗;η], Φ δ(0,0 ) = δJ [y∗;ζ],

(26)

and

Ψ 𝜖(0,0) = δK [y∗;η], Ψ δ(0, 0) = δK [y∗;ζ ].

(27)

Define the constant

δJ-[y∗;ζ] λ = − δK [y∗;ζ].

(28)

Then

|--------------------------| |δJ [y ;η] + λδK [y ;η] = 0 | -----∗------------∗--------

(29)

for every endpoint-preserving $η$ . This proves the theorem. □

4 From the multiplier theorem to Euler–Lagrange

The first variations are

∫ b ′ δJ = (Fy η + Fy′η )dx a

(30)

and

∫ b ′ δK = (Gy η + Gy′η )dx. a

(31)

Therefore

∫ b δJ + λδK = [(Fy + λGy )η + (Fy ′ + λGy ′)η′]dx. a

(32)

Introduce the augmented integrand

|---------′-----------′-------------′--| -H-(x,y,y-)-=-F-(x,y,y-) +-λG-(x,-y,y-).-|

(33)

Then

∫ b δJ + λ δK = δ H dx. a

(34)

Thus the constrained extremal must satisfy the ordinary Euler–Lagrange equation for $H$ :

|----------------| | -d- | Hy − dx Hy ′ = 0. ------------------

(35)

Equivalently,

|--------------------------------| |F + λG − d--(F ′ + λG ′) = 0.| --y------y---dx---y------y--------

(36)

The original integral constraint must still be imposed after solving this differential equation. It is normally used to determine the multiplier $λ$ or another remaining constant.

Figure. For a normal integral constraint, augment the integrand by a constant multiplier, derive the Euler–Lagrange equation for the augmented integrand, apply the endpoint conditions, and finally enforce the original integral constraint.

5 Why the multiplier is a constant

The condition

K [y] = k0

(37)

is one scalar equation on the entire function. Its multiplier is therefore one scalar constant $λ$ .

This should be distinguished from a pointwise constraint such as

C (x,y(x),z(x )) = 0 for every x.

(38)

A pointwise constraint represents infinitely many scalar conditions, one at each $x$ , and its multiplier is generally a function $λ(x)$ . That more mechanical constraint structure will be developed later in the series.

6 Worked example: minimum gradient energy with prescribed mean

Consider

∫ 1 J[y] = 1- (y′)2 dx 2 0

(39)

with fixed endpoints

y(0) = y(1) = 0

(40)

and the integral constraint

∫ 1 y(x) dx = m. 0

(41)

Here

1 ′2 F = --(y ) , G = y. 2

(42)

The augmented integrand is

1 H = --(y′)2 + λy. 2

(43)

Its derivatives are

Hy = λ, Hy′ = y′.

(44)

The Euler–Lagrange equation is

′′ λ − y = 0,

(45)

′′ y = λ.

(46)

Integrating twice gives

λ- 2 y(x) = 2 x + C1x + C2.

(47)

The endpoint conditions imply

λ C2 = 0, C1 = − --, 2

(48)

y(x) = − λ-x(1 − x). 2

(49)

Now impose the integral constraint:

m	= − ∫ ₀¹x(1 − x) dx	(50)
	= −.	(51)

Therefore

λ = − 12m

(52)

and

|--------------------| y ∗(x ) = 6m x(1 − x).| ----------------------

(53)

Figure. The fixed-mean constraint forces the minimum-gradient-energy curve away from the zero function. The stationary curve is a parabola whose amplitude is set by the prescribed integral $m$ .

This example can be classified directly. Let

y = y + u, ∗

(54)

where

∫ 1 u(0) = u(1) = 0, u dx = 0. 0

(55)

Then

J[y]	= ∫ ₀¹(y_∗′ + u′)²dx	(56)
	= J[y_∗] + ∫ ₀¹y_∗′u′dx + ∫ ₀¹(u′)²dx.	(57)

Integrating the cross term by parts,

∫ ∫ ∫ 1 ′ ′ 1 ′′ 1 y ∗u dx = − y∗ u dx = − λ u dx = 0. 0 0 0

(58)

Hence

|----------------∫---------------| | 1- 1 ′2 | |J[y] − J[y∗] = 2 (u ) dx ≥ 0. | -------------------0-------------

(59)

Equality requires $u′ = 0$ , and the endpoint conditions then force $u = 0$ . Thus $y ∗$ is the unique global minimizer, not merely a stationary function.

7 Worked example: normalization constraint and an eigenvalue problem

Consider

∫ π J[y] = (y′)2dx 0

(60)

with

y(0) = y(π) = 0

(61)

and normalization

∫ π y2dx = 1. 0

(62)

Use

H = (y ′)2 + λy2.

(63)

Then

H = 2λy, H ′ = 2y ′, y y

(64)

so Euler–Lagrange gives

y ′′ = λy.

(65)

Nontrivial solutions satisfying both endpoint conditions require

λ = − n2, n = 1,2,3,...,

(66)

with

y (x) = A sin(nx ). n n

(67)

Normalization gives

∘ -- 2 An = -- π

(68)

up to an overall sign. Thus

|--------∘-----------| | 2 | yn (x ) = --sin(nx).| -----------π----------

(69)

For these stationary functions,

2 J [yn] = n .

(70)

Therefore the lowest branch is

|--------∘-----------------------| | 2- | |y1(x) = π sinx, J [y1] = 1.| ----------------------------------

(71)

Figure. The normalization constraint converts the multiplier into an eigenvalue parameter. The stationary functions form discrete sine modes, and the lowest mode has the smallest gradient energy. This is the elementary variational structure behind later Rayleigh–Ritz and Sturm–Liouville theory.

The sign assigned to the multiplier is conventional. If the augmented functional is written $J − μK$ rather than $J + λK$ , then $μ = − λ = n2$ is positive.

8 The classical isoperimetric geometry preview

The name isoperimetric comes from geometric problems in which one quantity, historically perimeter or length, is fixed while another quantity such as enclosed area is extremized. A graph version illustrates the multiplier method without yet proving the full closed-curve isoperimetric theorem.

Suppose one seeks a stationary graph for the area functional

∫ b A [y] = ydx a

(72)

subject to fixed arc length

∫ b∘ --------- L[y] = 1 + (y′)2dx = L . a 0

(73)

The augmented integrand is

H = y + λ∘1--+-(y′)2.

(74)

Euler–Lagrange gives

( ′ ) 1 − d-- λ∘----y------ = 0. dx 1 + (y′)2

(75)

Since $λ$ is constant,

y′′ 1 ------′-2-3∕2-= -. (1 + (y ) ) λ

(76)

The left-hand side is the signed curvature of the graph. Therefore every normal stationary arc has constant curvature. A nonzero constant-curvature plane curve is a circular arc. The full isoperimetric theorem requires additional global arguments; CV09E1 will work through the classical geometry in more detail.

9 Multiple integral constraints

Suppose there are $m$ scalar constraints

∫ b Kr [y] = Gr (x,y,y′)dx = kr, r = 1,...,m. a

(77)

Under the appropriate independence condition, introduce constants

λ1,...,λm

(78)

and form

∑m H = F + λrGr. r=1

(79)

Then a normal constrained extremal satisfies

|----------------| | -d- | |Hy − dx Hy ′ = 0| ------------------

(80)

along with all original constraints and endpoint conditions.

The multiplier vector is determined together with the integration constants. Linear dependence among the first variations of the constraints requires special care because the effective number of independent constraints may be smaller than the number written down.

10 Normal and abnormal extremals

The theorem above assumed that the first variation of the constraint is not the zero functional. This is the normal case.

δK [y∗;η] = 0

(81)

for every admissible $η$ , then the constraint itself is stationary at $y ∗$ . The proof based on solving the constraint for a correction parameter breaks down because the required derivative $Ψ δ$ vanishes.

A more general multiplier statement introduces two constants $λ0$ and $λ1$ , not both zero, such that

λ δJ + λ δK = 0. 0 1

(82)

If $λ0 ⁄= 0$ , rescaling gives the normal form used throughout this article. If $λ0 = 0$ , the extremal is called abnormal. Abnormality is important in advanced variational theory and optimal control, but most elementary isoperimetric examples are normal. See [1, 2] for broader formulations.

11 What the multiplier means

The multiplier has several useful interpretations.

First, it is the coefficient that converts the constrained stationarity problem into unconstrained stationarity of an augmented functional.

Second, in regular optimization problems it measures the first-order sensitivity of the optimal value to the prescribed constraint value. Schematically, if

Jmin = Jmin(k0),

(83)

then, subject to sign convention and regularity,

dJmin- λ ∼ − dk0 .

(84)

This is the infinite-dimensional analogue of the shadow price interpretation of an ordinary Lagrange multiplier.

Third, in eigenvalue problems the multiplier often becomes the spectral parameter. The normalization constraint prevents the trivial zero function, and stationarity selects discrete eigenfunctions and eigenvalues.

12 Common mistakes

Varying freely while ignoring the constraint. An arbitrary perturbation normally leaves the constraint surface.
Requiring $δK = 0$ for every test function and then applying the Fundamental Lemma directly. Only tangent directions satisfy the linearized constraint; the multiplier theorem restores arbitrary test directions through the augmented functional.
Forgetting to impose the original constraint after solving Euler–Lagrange. The augmented differential equation alone does not determine the multiplier.
Treating $λ$ as a function of $x$ for a single scalar integral constraint. The multiplier is a constant because the constraint is one scalar equation.
Confusing an integral constraint with a pointwise constraint. Pointwise constraints generally require multiplier functions.
Assuming the multiplier condition proves a minimum. It gives a necessary stationarity condition in the normal case. Classification still requires additional arguments.
Forgetting multiplier sign conventions. Writing $F + λG$ or $F − μG$ is equivalent after redefining the multiplier.
Ignoring abnormal extremals. The simple theorem requires a nondegenerate constraint first variation.

13 A compact workflow

For one scalar integral constraint, use the following sequence:

Write the objective functional $J[y]$ .
Write the constraint functional $K [y] = k0$ .
Verify the endpoint conditions and identify admissible variations.
Check that the constraint is normal, at least at the classical level.
Introduce a constant multiplier $λ$ .
Form the augmented integrand $H = F + λG$ .
Apply the ordinary Euler–Lagrange equation to $H$ .
Apply all endpoint conditions.
Impose the original integral constraint to determine $λ$ and remaining constants.
Separately decide whether the stationary function is a minimum, maximum, or saddle.

14 What CV10 adds

CV09 keeps each extremal smooth across the whole interval but restricts which functions are admissible through integral constraints. CV10 changes a different assumption: the stationary path may be only piecewise smooth and may contain an interior corner whose location can move.

That leads to the Weierstrass–Erdmann corner conditions, which determine which generalized momentum and Hamiltonian-like quantities must remain continuous across a movable corner.

Summary

For

∫ b ′ J [y] = F(x,y, y) dx a

(85)

subject to the scalar integral constraint

∫ b K [y] = G (x,y,y′)dx = k0, a

(86)

a normal constrained extremal satisfies

|--------------| -δJ-+-λδK--=-0--

(87)

for some constant $λ$ . Equivalently, with

H = F + λG,

(88)

the extremal satisfies

|----------------| Hy − -d-Hy ′ = 0. ------dx----------

(89)

The multiplier theorem is not merely an algebraic trick. It is the statement that, at a constrained extremum, the first variation of the objective is linearly dependent on the first variation of the constraint. This idea reappears in isoperimetric geometry, eigenvalue theory, constrained mechanics, field theory, and optimal control.

References

[1] I. M. Gelfand and S. V. Fomin, Calculus of Variations, Dover Publications, 2000.

[2] Bruce van Brunt, The Calculus of Variations, Springer, 2004.

[3] Hans Sagan, Introduction to the Calculus of Variations, Dover Publications, 1992.

[4] Gilbert Ames Bliss, Lectures on the Calculus of Variations, University of Chicago Press, 1946.

[5] Richard Courant and David Hilbert, Methods of Mathematical Physics, Volume I, Wiley-Interscience, 1989 reprint.

"Calculus of Variations: Constrained Variations and Isoperimetric Problems" is owned by bloftin.
(view preamble)