Calculus of Variations: Beltrami Problems and Catenary-Type Examples

CV06 derived the Beltrami identity for an autonomous first-order functional

(1)

namely

(2)

The importance of this identity is computational: it replaces the usual second-order Euler–Lagrange equation by a first-order relation. In favorable cases that relation can be solved directly for y′ and then integrated by a single quadrature. This companion set develops that skill through increasingly physical examples, culminating in the catenary and catenoid and ending with a preview of the brachistochrone [1, 2, 3].

1 How to use this set

Attempt every exercise before reading Part II. For each problem, use the same sequence:

1. identify the integrand F and verify that F_x = 0;
2. compute F_y′ carefully;
3. form F − y′F_y′ before simplifying;
4. set the result equal to a constant;
5. solve the first-order relation for y′ or dx∕dy;
6. integrate once more, keeping track of branches and constants;
7. apply endpoint or symmetry data only after the general stationary family has been obtained; and
8. remember that satisfying Beltrami proves stationarity under the stated hypotheses, not automatically global minimality.

Part I: Exercises

Exercise 1: recognize when Beltrami applies

For each integrand below, state whether the Beltrami identity can be used immediately. If it can, compute the corresponding first integral.

1. F = .
2. F = (y′)² + V (y).
3. F = x²(y′)² + y².
4. F = e^y.
5. F = (1 + x)(y′)².

For the cases where Beltrami does not apply, identify the explicit dependence that prevents its direct use.

Exercise 2: a master weighted-length family

Consider

(3)

1. Use Beltrami to show that every sufficiently smooth stationary curve satisfies

   (4)

2. Solve for (y′)².
3. Derive the quadrature

   (5)

4. Explain why the condition Φ(y) ≥|C| appears automatically.

This exercise is the common algebraic core of several later examples.

Exercise 3: shortest path in the plane revisited

For the arc-length functional

(6)

use Beltrami, rather than the full Euler–Lagrange equation, to show that every smooth stationary graph has constant slope and is therefore a straight line. Then impose

(7)

and obtain the explicit stationary curve.

Exercise 4: the catenary from an augmented chain functional

A uniform flexible chain in a vertical plane has gravitational potential energy proportional to

(8)

Its total length is fixed. Introducing a constant multiplier λ for that length constraint gives the augmented integrand

(9)

For this exercise, take the augmented functional as given; CV09 will derive the variational multiplier rule systematically.

1. Apply Beltrami and show that

   (10)

   where a > 0 is a constant.

2. Solve for y′ and separate variables.
3. Integrate to obtain

   (11)

4. Show that the lowest point occurs at x = b and has horizontal tangent.
5. Explain which constants are fixed by geometry and which constant is associated with the length constraint.

Exercise 5: why a shallow catenary looks parabolic

For the symmetric catenary

(12)

use the Taylor expansion of cosh z to show that, for |x|≪ a,

(13)

Hence derive the leading parabolic approximation

(14)

Estimate the first neglected correction term and explain why a hanging cable with small sag can appear almost parabolic even though its exact ideal-chain shape is a catenary.

Exercise 6: the catenoid from minimum surface area

A surface of revolution is formed by rotating the graph y(x) > 0 about the x-axis. Its area is

(15)

1. Ignore the constant factor 2π and apply Beltrami.
2. Show that

   (16)

3. Integrate the first-order equation and obtain

   (17)

4. Explain why the same hyperbolic cosine appears in both the hanging chain and minimum-surface problems even though the physical functionals are different.

Exercise 7: autonomous mechanics and the energy integral

Let the independent variable be time t and write q_t = dq∕dt. Consider the action

(18)

1. Apply the Beltrami identity with F = L(q,q_t).
2. Show that the result can be written

   (19)

3. Solve for dt∕dq and obtain the quadrature

   (20)

4. Explain why turning points satisfy V (q) = E.

Exercise 8: brachistochrone preview by Beltrami

Let y measure vertical distance downward from the starting point. Conservation of mechanical energy gives speed

(21)

The travel time along a graph y(x) is therefore

(22)

1. Apply Beltrami and show that the first integral is equivalent to

   (23)

   for some positive constant a.

2. Introduce the parameter

   (24)

   and show that

   (25)

3. Derive

   (26)

   which is a cycloid.

4. Explain why this exercise is only a preview: CV17 will address the historical problem, endpoint geometry, and full interpretation in detail.

Part II: Complete Worked Solutions

Solution 1: recognize when Beltrami applies

Beltrami requires that the integrand have no explicit dependence on the independent variable x.

(a) For

(27)

we have F_x = 0. Also

(28)

Thus

F − y′Fy′	= −	(29)
	= = C.	(30)

(b) For

(31)

again F_x = 0. Since F_y′ = y′,

(32)

(c) The integrand

(33)

contains x explicitly, so Beltrami does not apply directly.

(d) For

(34)

there is no explicit x dependence. Therefore

(35)

(e) The factor (1 + x) is explicit x dependence, so the Beltrami identity cannot be replaced by a constant. The more general du Bois–Reymond identity from CV06 must be used instead.

Solution 2: a master weighted-length family

Let

(36)

Because F_x = 0, Beltrami gives

(37)

First compute

(38)

Therefore

F − y′Fy′	= Φ(y) − Φ(y)	(39)
	= .	(40)

Hence

(41)

Squaring gives

(42)

so

(43)

On an interval where a consistent branch is chosen,

(44)

Integrating,

(45)

For the square root to remain real,

(46)

or

(47)

because Φ > 0. This restriction is not imposed separately; it is encoded in the first integral itself.

Solution 3: shortest path in the plane revisited

Here

(48)

From Solution 1,

(49)

Therefore

(50)

which is constant. Choosing one continuous branch,

(51)

where m is constant. Hence

(52)

Apply the two endpoint conditions:

(53)

Subtracting,

(54)

Thus

(55)

Beltrami has recovered the same stationary line as the full Euler–Lagrange calculation, but with one fewer differentiation step.

Solution 4: the catenary from an augmented chain functional

Take

(56)

There is no explicit x dependence. Compute

(57)

Then

F − y′Fy′	= (y + λ)	(58)
	= .	(59)

Set the constant equal to a positive parameter a:

(60)

Rearrange:

(61)

so

(62)

Invert the derivative on a monotone branch:

(63)

Let

(64)

Then

(65)

Because cosh is even, both branches combine into

(66)

Therefore

(67)

Differentiate:

(68)

Thus

(69)

and since cosh z ≥ 1, the point x = b is the lowest point of the curve for this sign convention.

The constant b locates the horizontal position of the lowest point. The constant a sets the curvature scale. The multiplier λ enters as a vertical shift in this augmented form and is ultimately determined together with a and b by the endpoint geometry and the prescribed total chain length. The justification for introducing λ as a variational multiplier is the subject of CV09 [1, 2].

Solution 5: why a shallow catenary looks parabolic

Use

(70)

With z = x∕a,

y(x)	= a	(71)
	= a	(72)
	= + + .	(73)

Therefore, when |x|∕a ≪ 1,

(74)

The leading neglected term is

(75)

Relative to the quadratic term, its size is approximately

(76)

Thus if |x|∕a = 0.3, for example, the first correction is only about

(77)

or less than one percent of the quadratic term. This is why shallow catenaries are visually difficult to distinguish from parabolas over a limited span.

Solution 6: the catenoid from minimum surface area

Ignoring the constant factor 2π, the integrand is

(78)

Beltrami gives

(79)

Since

(80)

we obtain

(81)

where a > 0. Rearranging,

(82)

Thus

(83)

Integrating exactly as in the catenary calculation gives

(84)

The repeated hyperbolic cosine is not an accident. Both problems reduce to an integrand of the general weighted-length form

(85)

with a weight linear in y after a vertical shift. The physical meanings are different: the catenary comes from gravitational potential energy with a length constraint, while the catenoid comes from surface area. The algebraic structure of the Beltrami first integral is nevertheless the same [3, 2].

Solution 7: autonomous mechanics and the energy integral

Let

(86)

Because L has no explicit time dependence, Beltrami gives

(87)

Now

(88)

Therefore

L − qt Lqt	= m(qt)2 − V (q) − m(q t)2	(89)
	= −.	(90)

Writing C = −E gives

(91)

Solve for q_t:

(92)

Invert:

(93)

Hence

(94)

At a turning point, q_t = 0. Therefore the energy equation requires

(95)

This is the familiar mechanical energy integral, obtained here as a direct Beltrami first integral of an autonomous action [4, 5].

Solution 8: brachistochrone preview by Beltrami

The integrand is

(96)

There is no explicit x dependence. Compute

(97)

Then

F − y′Fy′	=	(98)
	= = C.	(99)

Square and absorb the positive constants into a new parameter a:

(100)

Now set

(101)

The first integral gives

(102)

Therefore

(103)

and on the descending branch

(104)

Differentiate the parameterization of y:

(105)

Since

(106)

we have

	=	(107)
	= a sin 𝜃 tan	(108)
	= a(1 − cos 𝜃).	(109)

Integrating,

(110)

Together with

(111)

this is the parametric equation of a cycloid. The appearance of the cycloid is one of the classical achievements of the early calculus of variations. CV17 will return to the brachistochrone with its historical development and complete variational interpretation [1, 3].

What these problems should teach

The most important lesson is not that several famous curves can be memorized. It is that one structural observation,

(112)

changes the solution strategy. Rather than expand the full Euler–Lagrange equation into a second-order ODE, first form

(113)

For weighted-length integrands of the form

(114)

this immediately becomes

(115)

which often exposes the geometry of the problem before any difficult integration begins.

Common mistakes

• Using Beltrami when F contains x explicitly. The constant first integral requires F_x = 0.
• Computing F_y′ incorrectly. In Φ(y), the factor Φ(y) is held constant when taking the partial derivative with respect to y′.
• Dropping the square-root domain condition. Solving for y′ can introduce a requirement such as Φ(y)² ≥ C².
• Forgetting the branch sign. A first-order relation usually gives y′ = ±f(y). A complete smooth curve may switch monotone branches at a turning point.
• Treating the multiplier λ in the catenary as arbitrary decoration. It enforces the fixed-length constraint; CV09 provides the theorem justifying it.
• Calling every hyperbolic-cosine graph a hanging chain. The same analytic profile also generates a catenoid, but the underlying variational functional is different.
• Assuming a first integral proves a minimum. Beltrami supplies a necessary stationarity relation. Classification is a separate issue.

Summary

For an autonomous functional

(116)

stationarity implies the Beltrami identity

(117)

For the common weighted-length family

(118)

this reduces to

(119)

The same algebraic structure produces a straight line for ordinary planar arc length, a hyperbolic cosine for the catenary and catenoid, an energy integral in autonomous mechanics, and the first-order cycloidal relation in the brachistochrone. CV06E2 next emphasizes the complementary special case of cyclic dependent variables and conserved conjugate momenta.

References