Calculus of Variations: Constrained Physical Systems - Worked Examples

(Example)

Calculus of Variations: Constrained Physical Systems - Worked Examples and Applications

CV09 developed the variational Lagrange-multiplier method for an integral constraint. CV09E1 applied that machinery to the classical isoperimetric problem. This companion article turns to physical systems in which the constraint has a direct mechanical, vibrational, or quantum interpretation.

The central pattern is

∫ b ′ ∫ b ′ J[y] = F (x,y,y )dx, K [y] = G (x,y,y )dx = k0, a a

(1)

followed by the augmented functional

∫ ^ b J [y] = (F + λG )dx, a

(2)

where the multiplier λ is constant for a single scalar integral constraint. In physical applications, that multiplier often has an additional interpretation: it may act like a reaction load, a horizontal Tension scale, a squared natural frequency, an energy eigenvalue, or a generalized force associated with the imposed global condition. This is one reason constrained variational methods are so useful across mechanics and physics [1, 2, 3].

Figure. The same constrained-variation template appears in apparently different physical systems. The multiplier is introduced mathematically to enforce the integral constraint; its physical interpretation depends on the problem.

1 Exercises

Exercise 1: taut string with prescribed mean deflection

A taut string on 0 ≤ x ≤ L has constant tension T and fixed endpoints

y (0) = y(L) = 0.

(3)

Its small-slope elastic energy is

∫ L T- ′ 2 U [y] = 2 0 (y ) dx.

(4)

Among all admissible shapes satisfying the prescribed area/mean-deflection condition

∫ L y (x )dx = A, 0

(5)

find the stationary shape and determine the multiplier. Then evaluate the minimum energy.

Exercise 2: string on an elastic foundation with prescribed mean deflection

Let

∫ [ ] L T- ′2 k- 2 U[y] = 2 (y ) + 2 y dx, 0

(6)

with T > 0, k > 0, and y(0) = y(L) = 0. Impose

∫ L ydx = A. 0

(7)

Derive the constrained Euler–Lagrange equation.
Solve it in a form symmetric about x = L∕2.
Determine the multiplier from the integral constraint.
Show that the result tends to the parabola of Exercise 1 as k → 0.

Exercise 3: the hanging chain as a fixed-length constrained system

A uniform flexible chain with Weight per unit arc length w hangs in a uniform gravitational field. Its endpoints are fixed horizontally, and its total length is prescribed:

∫ x2∘ --------- 1 + (y′)2dx = ℓ. x1

(8)

Take the vertical coordinate y positive upward. The gravitational potential energy is

∫ x 2 ∘ ------′-2 U [y] = w x y 1 + (y ) dx. 1

(9)

Form the augmented integrand.
Use the Beltrami identity to obtain a first integral.
Show that the stationary curve is a catenary
$( ) y(x) = acosh x-−-x0- − c. a$ (10)
Interpret the constant wa physically.

Exercise 4: normalized vibration mode of a uniform string

A uniform string of mass density μ and tension T is fixed at x = 0 and x = L. Minimize the strain-energy functional

T ∫ L ′2 U [ϕ] = -- (ϕ ) dx 2 0

(11)

subject to the normalization

∫ L μ ϕ2dx = 1 0

(12)

and endpoint conditions ϕ(0) = ϕ(L) = 0.

Derive the eigenvalue equation.
Find the normalized stationary functions.
Identify the multiplier with the squared natural frequency.
Show which stationary function minimizes U under the normalization constraint.

Exercise 5: variable-density string as a Sturm–Liouville problem

Let T(x) > 0 and μ(x) > 0. Minimize

1∫ L U [ϕ ] = -- T (x)(ϕ′)2 dx 2 0

(13)

subject to

∫ L 2 μ(x)ϕ dx = 1, ϕ(0) = ϕ(L ) = 0. 0

(14)

Derive the differential equation and explain why the multiplier is again an eigenvalue parameter.

Exercise 6: normalization in the one-dimensional infinite square well

For a real wavefunction ψ(x) on 0 ≤ x ≤ L, consider the kinetic-energy expectation functional

∫ L 2 E [ψ ] = ℏ--(ψ′)2dx, 0 2m

(15)

with

∫ L 2 ψ(0) = ψ (L) = 0, ψ dx = 1. 0

(16)

Introduce a multiplier and derive the stationary equation.
Find all normalized stationary states.
Show that the multiplier equals the energy expectation at a stationary state.
Identify the minimum-energy state.

Exercise 7: a particle trajectory with a prescribed time integral of position

A particle moves on a line during 0 ≤ t ≤ τ. Consider

m ∫ τ 2 J[x] = -- ˙x dt 2 0

(17)

with

x(0) = x(τ) = 0.

(18)

Impose the global trajectory constraint

∫ τ x(t)dt = A. 0

(19)

Find the stationary trajectory and interpret the multiplier as an effective constant generalized force.

Exercise 8: selecting the second string mode with two constraints

For the same uniform string as Exercise 4, minimize

∫ L U [ϕ] = T- (ϕ ′)2dx 2 0

(20)

subject to both

∫ L 2 μ ϕ dx = 1 0

(21)

and orthogonality to the normalized fundamental mode ϕ₁:

∫ L μ ϕϕ1 dx = 0. 0

(22)

Use two multipliers to derive the stationarity equation and show that the minimizer is the normalized second mode.

2 Solutions

Solution 1: taut string with prescribed mean deflection

The unconstrained integrand is

T- ′2 F = 2 (y ) ,

(23)

and the constraint integrand is simply

G = y.

(24)

Use

H = T-(y′)2 + λy. 2

(25)

Then

Hy = λ, Hy′ = T y′,

(26)

so Euler–Lagrange gives

λ − T y′′ = 0.

(27)

Thus

′′ λ- y = T .

(28)

Integrating twice,

λ-- 2 y(x) = 2T x + C1x + C2.

(29)

The fixed endpoints give C₂ = 0 and

λL C1 = − ---, 2T

(30)

hence

y(x) = -λ-x(x − L ). 2T

(31)

Now impose the global constraint:

A	= ∫ ₀^Lx(x − L) dx	(32)
	=	(33)
	= −.	(34)

Therefore

|------------| | 12T A | |λ = − ---3--| --------L-----

(35)

and

|--------------------| y∗(x ) = 6A-x(L − x).| ---------L3-----------

(36)

For A > 0, the multiplier is negative in the convention H = F + λG.

Figure. The minimum-energy string shape under a prescribed integral of displacement is a parabola. The global constraint fixes its amplitude.

The derivative is

y ′(x ) = 6A-(L − 2x). ∗ L3

(37)

Therefore

U[y_∗]	= ∫ ₀^L(L − 2x)² dx	(38)
	=	(39)
	= .	(40)

Because the functional is strictly convex in y′ and the constraint is affine, this stationary function is the unique minimizer within the fixed-endpoint admissible class.

Solution 2: elastic foundation plus prescribed mean deflection

The augmented integrand is

T k H = --(y′)2 + --y2 + λy. 2 2

(41)

Euler–Lagrange gives

ky + λ − T y′′ = 0,

(42)

Ty′′ − ky = λ.

(43)

Define

∘ --- k βL β = --, α = ---. T 2

(44)

A form that makes the symmetry about L∕2 explicit is

[ ] λ- cosh-(β-(x-−-L∕2-)) y(x) = k coshα − 1 .

(45)

At x = 0 and x = L, the ratio of coshes is one, so the endpoint conditions hold automatically.

Integrating,

A	=	(46)
	= .	(47)

Hence

|--------------------| | -----kA------ | |λ = − 2 .| | L − --tanh α | -----------β----------

(48)

The denominator is positive, so A > 0 again corresponds to λ < 0.

To recover the k → 0 limit, use

z2- 4 α3- 5 cosh z = 1 + 2 + O (z ), tanh α = α − 3 + O (α ).

(49)

Since β² = k∕T, expansion of the solution and multiplier gives

y(x) −→ 6A-x(L − x), L3

(50)

which is precisely Exercise 1.

Solution 3: hanging chain with fixed length

The constrained problem is to minimize gravitational potential energy while keeping the total arc length fixed. The augmented integrand is

∘ --------- H = (wy + λ) 1 + (y′)2.

(51)

There is no explicit x dependence, so the Beltrami identity gives

H − y′Hy′ = C.

(52)

Now

----y′----- Hy ′ = (wy + λ) ∘1-+--(y′)2,

(53)

and therefore

H − y′H_y′	= (wy + λ)	(54)
	= .	(55)

Thus

|----------------| | wy + λ | |∘----------= C. | ---1-+-(y′)2-------

(56)

Define

a = C-, c = -λ. w w

(57)

Then

---y +-c--- ∘ ------′-2 = a. 1 + (y )

(58)

Rearranging,

′2 (y-+-c)2 (y) = a2 − 1.

(59)

The standard integration gives

|-------------(-------)------| |y(x) = acosh x-−-x0- − c.| | a | ------------------------------

(60)

Figure. The fixed-length constraint introduces a multiplier into the potential-energy problem. The Beltrami first integral reduces the resulting equation to the catenary.

For a static chain, the horizontal component of the tension is constant. Denoting it by H₀, elementary force balance gives the familiar catenary scale

H0 = wa.

(61)

But wa = C from the first integral, so the Beltrami constant is precisely the constant horizontal tension. This is a useful example in which the variational constant has an immediate mechanical interpretation. The catenary and its variational formulation are classical applications of the calculus of variations [2, 1].

Solution 4: normalized vibration mode of a uniform string

Use the augmented functional

∫ [ ] ^ L T- ′ 2 λ- 2 U[ϕ] = 2(ϕ ) − 2μ ϕ dx. 0

(62)

The factor 1∕2 is merely convenient. Euler–Lagrange gives

′′ − λμϕ − Tϕ = 0,

(63)

|-----′′--------| -−-Tϕ--=--λμϕ.-|

(64)

For constant T and μ,

λμ ϕ′′ + k2ϕ = 0, k2 = --. T

(65)

The fixed endpoints require

n π kn = ---, n = 1,2,3, ... L

(66)

and hence

( ) λ = T- n-π 2. n μ L

(67)

The normalization gives

∫ L ( nπx ) L 1 = μA2n sin2 ---- dx = μA2n--, 0 L 2

(68)

|--------∘-----------------| | 2 ( nπx ) | |ϕn(x) = ---sin ---- .| ------------μL-------L-----

(69)

For the physical wave equation

μytt = T yxx,

(70)

Separation of variables gives

′′ 2 − T ϕ = μω ϕ.

(71)

Therefore

|------2-| λn-=--ωn.-

(72)

Figure. Normalization removes the arbitrary amplitude of each eigenmode. Among all normalized endpoint-zero functions, the fundamental mode has the smallest strain energy.

Multiply the eigenvalue equation by ϕ_n and integrate. Integration by parts gives

∫ L ∫ L T (ϕ ′)2 dx = λ μϕ2 dx = λ . 0 n n 0 n n

(73)

Hence

λn U[ϕn] = ---. 2

(74)

Because λ_n ∝ n², the minimum occurs at n = 1. This variational characterization of vibration modes is a prototype of the Rayleigh quotient and more general eigenvalue principles [4, 5].

Solution 5: variable-density string

Use

1 λ H = -T (x)(ϕ′)2 − -μ (x )ϕ2. 2 2

(75)

Then

H ϕ = − λ μ(x)ϕ, H ϕ′ = T (x)ϕ′.

(76)

Therefore

d-- ′ − λμ(x )ϕ − dx [T (x)ϕ ] = 0,

(77)

------------------------------- | d | |− ---[T(x)ϕ ′(x)] = λμ(x)ϕ (x).| ---dx--------------------------|

(78)

This is a Sturm–Liouville eigenvalue problem with weight μ(x). In the vibration interpretation, λ = ω². The normalization fixes the eigenfunction scale, while the endpoint conditions and differential equation select the admissible eigenvalues.

Solution 6: normalization in the infinite square well

The constrained functional is

∫ L [ ℏ2 ] ^E[ψ ] = ----(ψ′)2 − λ ψ2 dx. 0 2m

(79)

Euler–Lagrange gives

2 − 2λ ψ − ℏ--ψ′′ = 0, m

(80)

|--------------| | 2 | − -ℏ--ψ′′ = λ ψ. --2m------------

(81)

This is the time-independent Schrödinger equation inside an infinite square well, with the variational multiplier playing the role of the energy eigenvalue. The boundary conditions give

( nπx-) ψn (x) = An sin L ,

(82)

and normalization gives

|------------------------| | ∘ 2 (n πx) | ψn (x) = -- sin ---- .| -----------L--------L-----

(83)

The corresponding multipliers are

|--------------| | ℏ2π2n2 | |λn = 2mL2---.| ---------------

(84)

To verify that the multiplier equals the energy functional, multiply the differential equation by ψ_n and integrate:

2 ∫ L ∫ L − -ℏ-- ψ ψ ′′dx = λ ψ2 dx. 2m 0 n n n 0 n

(85)

The normalization makes the right side equal to λ_n. Integrating the left side by parts and using ψ_n(0) = ψ_n(L) = 0 gives

ℏ2 ∫ L ---- (ψ ′n)2dx = λn. 2m 0

(86)

Thus

|------------| |E [ψn ] = λn.| -------------

(87)

The minimum occurs at n = 1. This is the simplest example of the quantum variational principle with a normalization constraint [6].

Solution 7: constrained particle trajectory

The augmented integrand is

m 2 H = --x˙ + λx. 2

(88)

Euler–Lagrange gives

λ − m ¨x = 0,

(89)

-λ ¨x = m .

(90)

With x(0) = x(τ) = 0,

x (t) = -λ--t(t − τ). 2m

(91)

The integral constraint gives

A	= ∫ ₀^τt(t − τ) dt	(92)
	= −.	(93)

Hence

|------------| | 12mA--| λ = − τ3 | --------------

(94)

and

|--------------------| |x (t) = 6A-t(τ − t). | --∗------τ-3---------|

(95)

Figure. Without the integral constraint the kinetic functional is minimized by the zero path. Prescribing the time integral of position forces a parabolic trajectory.

The equation

m ¨x = λ

(96)

has the form of Newton’s law under a constant generalized force. In the convention used here, positive prescribed A requires λ < 0 because the augmented term was chosen as +λx. Reversing the sign convention for the multiplier reverses that sign but not the physical trajectory.

Solution 8: selecting the second string mode with two constraints

Introduce two multipliers and use

T ′ 2 λ 2 H = --(ϕ ) − -μϕ − γ μϕϕ1. 2 2

(97)

Euler–Lagrange gives

|---------------------| −-T-ϕ′′ =-λμ-ϕ +-γμϕ1.--

(98)

To determine γ, multiply by ϕ₁ and integrate. Using integration by parts and the fundamental-mode equation

− Tϕ ′′1 = λ1μϕ1,

(99)

we obtain

T ∫ ₀^Lϕ′ϕ₁′dx	= λ∫ ₀^Lμϕϕ₁ dx + γ ∫ ₀^Lμϕ₁² dx	(100)
	= γ.	(101)

But another integration by parts gives

∫ ∫ L ′ ′ L T ϕ ϕ1 dx = λ1 μ ϕϕ1 dx = 0, 0 0

(102)

because of the orthogonality constraint. Therefore

|------| -γ-=-0.-

(103)

The stationarity equation reduces to the ordinary string eigenproblem

′′ − Tϕ = λμϕ,

(104)

but the extra constraint excludes the fundamental mode. Expanding any admissible normalized function in the orthonormal string modes,

∑∞ ∑∞ ϕ = c ϕ , c2 = 1, n n n n=2 n=2

(105)

its energy is

1 ∑∞ U[ϕ] = -- λnc2n. 2 n=2

(106)

Since

λ2 < λ3 < λ4 < ⋅⋅⋅ ,

(107)

the minimum is achieved by placing all weight in the second mode:

|----------------------| -ϕ-=-±-ϕ2,-----λ-=-λ2.-|

(108)

This is the basic constrained-minimization mechanism behind variational characterizations of higher eigenmodes: normalization fixes scale, while orthogonality constraints remove lower modes [5, 4].

3 What these physical examples teach

The same mathematical multiplier can play several different physical roles.

In the fixed-mean string problem, the multiplier acts like a uniform reaction load that enforces the prescribed total displacement.
In the hanging-chain problem, the Beltrami constant generated by the constrained functional equals the constant horizontal tension wa.
In vibration problems, the normalization multiplier becomes ω².
In the infinite square well, the normalization multiplier becomes the energy eigenvalue.
In the constrained trajectory problem, the multiplier appears in the differential equation exactly like a constant generalized force.

The unifying lesson is that a global integral condition changes the set of allowable variations. The multiplier restores the convenience of arbitrary local test functions by transferring the global restriction into the augmented integrand.

4 Common mistakes

Treating λ as a prescribed physical constant before solving the constraint. The multiplier is determined together with the stationary function by the original integral condition.
Forgetting units. The multiplier must have units that make F and λG compatible inside the augmented integrand.
Using a variable multiplier for a single scalar integral constraint. Here λ is constant; pointwise constraints require a multiplier field and belong to a different class of constrained problem.
Dropping the normalization after deriving the eigenvalue equation. The differential equation determines the shape only up to scale; the constraint fixes that scale.
Identifying every stationary eigenfunction with the minimum. Higher modes are stationary under normalization but have larger energy. Additional orthogonality constraints are needed to characterize them as constrained minima.
Confusing a fixed-length catenary problem with an unconstrained weighted-length functional without checking the multiplier. The catenary family can be written in several equivalent variational forms, but the physical fixed-length chain requires the length constraint.
Reading the sign of λ without checking the sign convention in F ± λG. The stationary curve is invariant under a simultaneous sign redefinition of the multiplier convention.

5 Bridge to later entries

CV09E2 already points toward several later parts of the series. The string and quantum examples anticipate the variational eigenvalue principles of CV21. The trajectory example connects to Hamilton’s principle in CV14 and optimal control in CV23. The catenary again shows the usefulness of the first-integral methods of CV06. Multiple orthogonality constraints foreshadow systematic constrained-mode and Rayleigh–Ritz calculations.

The next main entry, CV10, changes the kind of admissible curve rather than imposing a global integral constraint: stationary paths may be only piecewise smooth, leading to the Weierstrass–Erdmann corner conditions.

References

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

[2] Cornelius Lanczos, The Variational Principles of Mechanics, 4th ed., Dover Publications, 1986.

[3] Herbert Goldstein, Charles Poole, and John Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2001.

[4] Leonard Meirovitch, Elements of Vibration Analysis, 2nd ed., McGraw-Hill, 1986.

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

[6] David J. Griffiths and Darrell F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.

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Calculus of Variations: Constrained Physical Systems - Worked Examples