Calculus of Variations: Functional Evaluation and Comparison Exercises

This companion entry reinforces CV01 before any first-variation machinery is introduced. The emphasis is deliberately elementary but fundamental: a functional assigns a scalar to an entire admissible function, and the value of that scalar depends both on the formula for the functional and on the function inserted into it.

The exercises develop four habits that are essential later in the course:

1. identify the domain and admissible class before doing algebra;
2. substitute a complete trial function into a functional carefully;
3. compare several admissible candidates without assuming that stationarity machinery is required; and
4. recognize that a restricted family of functions can turn an infinite-dimensional problem into an ordinary finite-dimensional one.

1 Exercises

Exercise 1: function or functional?

For each mapping below, determine whether it is an ordinary function, a functional, or can be interpreted as either depending on its declared domain. For every functional, state a natural domain on which it is well defined.

1. f(x) = x² + 1.
2. J[y] = ∫ ₀¹y(x)² dx.
3. P[y] = y(1∕2).
4. L[y] = ∫ ₀¹ dx.
5. M[y] = max _0≤x≤1|y(x)|.

Explain why a functional need not be written as an integral.

Exercise 2: evaluate one functional on several functions

Consider

(1)

Evaluate J[y] for each of the following functions:

1. y₀(x) = 0,
2. y₁(x) = x,
3. y₂(x) = x²,
4. y₃(x) = sin(πx).

Rank the four functions by increasing functional value.

Exercise 3: the same formula, different variational problems

Consider the functional formula

(2)

Compare the following two admissible classes:

(3)

and

(4)

1. Determine whether y = x, y = x², and y = 0 belong to each class.
2. Evaluate J on every candidate for which it is admissible.
3. Explain why specifying only the formula for J does not completely specify the variational problem.
4. Without using Euler–Lagrange, identify the obvious global minimizer in 𝒜₂ and justify the claim.

Exercise 4: compare three admissible connecting curves

On the fixed-endpoint class

(5)

consider

(6)

and the three admissible curves

(7)

1. Verify the endpoint conditions.
2. Compute J[y₁], J[y₂], and J[y₃] exactly.
3. Which of these three trial curves has the smallest value?
4. Does winning this three-curve comparison prove that the winning curve is the global minimizer over all of 𝒜? Explain.

Exercise 5: arc length of a line and a parabola

For a graph y(x) on 0 ≤ x ≤ 1, define its length by

(8)

Both

(9)

connect (0, 0) to (1, 1).

1. Compute L[y₁] exactly.
2. Show that

   (10)

3. Compare the two values numerically and interpret the result geometrically.

Exercise 6: restrict the problem to a one-parameter family

For

(11)

every member satisfies y_c(0) = 0 and y_c(1) = 1.

For the functional

(12)

1. compute the ordinary scalar function J(c) = J[y_c];
2. minimize J(c) using ordinary single-variable calculus;
3. identify the corresponding curve;
4. explain what has and has not been proved about the full admissible class.

Exercise 7: a quadratic energy functional

Let

(13)

where T > 0 and K > 0 are constants. Evaluate E for the trial family

(14)

1. Find E(A) exactly.
2. Compare E[0], E[y_A], and E[y_2A].
3. Explain why the quadratic dependence on amplitude is typical of small-deflection elastic and field energies.

Exercise 8: evaluating a mechanical action on trial paths

Consider the one-dimensional action

(15)

with m > 0, k > 0, and fixed endpoint conditions q(0) = q(1) = 0. Restrict attention to the trial family

(16)

1. Compute S(a) = S[q_a] exactly.
2. For what relation between k and m is S(a) positive for every nonzero a in this trial family? When is it negative?
3. What happens when k = 10m?
4. Explain why this example warns against assuming that an “action principle” always means ordinary minimization of a positive quantity.

Exercise 9: an infimum that is not attained

Let

(17)

and define

(18)

1. Compute J[y_c].
2. Determine inf _y∈𝒜J[y].
3. Is that infimum attained by any admissible function?
4. What change to the admissible set would make the infimum attained?

2 Solutions

Solution 1: function or functional?

(a) The mapping

(19)

is an ordinary real-valued function of a numerical variable x.

(b) The mapping

(20)

is a functional because its input is an entire function y. A natural domain is C⁰([0, 1]), although much larger spaces such as L²(0, 1) are also possible.

(c) The map

(21)

is a functional if its input is a function y. A natural domain is C⁰([0, 1]). This example is important because it is a functional that is not written as an integral.

(d) The arc-length map

(22)

is a functional. A classical domain is C¹([0, 1]), possibly restricted by endpoint conditions.

(e) The map

(23)

is also a functional. For continuous functions on a compact interval the maximum exists, so C⁰([0, 1]) is a natural domain.

Thus a functional is characterized by the nature of its input, not by whether its formula contains an integral. Integrals are common because physical quantities such as length, action, energy, and mass accumulate local contributions, but point evaluation and maximum norms are also functionals.

Solution 2: evaluate one functional on several functions

The functional is

(24)

(a) For y₀ = 0, both y₀ and y₀′ vanish, so

(25)

(b) For y₁ = x, y₁′ = 1, hence

(26)

(c) For y₂ = x², y ₂′ = 2x, so

(27)

(d) For y₃ = sin(πx),

(28)

Therefore

J[y3]	= ∫ 01dx	(29)
	= + = .	(30)

Numerically,

(31)

Thus the ranking is

(32)

from smallest to largest functional value.

Solution 3: the same formula, different variational problems

The functional formula is

(33)

(a) For 𝒜₁, admissibility requires y(0) = 0 and y(1) = 1. Therefore

(34)

For 𝒜₂, admissibility requires both endpoints to vanish. Hence

(35)

while y = x and y = x² are not in 𝒜 ₂ because both equal 1 at x = 1.

(b) The admissible evaluations are

(36)

(37)

and

(38)

(c) The same integral expression can represent different optimization problems because the competitors are different. The admissible class is part of the mathematical definition of the problem.

(d) Since

(39)

we have

(40)

for every admissible y. The function y = 0 belongs to 𝒜₂ and attains J[0] = 0. Therefore it is a global minimizer. In fact it is the unique global minimizer among C¹ functions, because J[y] = 0 forces y′ = 0 everywhere, and the endpoint condition then forces y = 0.

Solution 4: compare three admissible connecting curves

(a) At x = 0,

(41)

At x = 1,

(42)

All three are admissible.

(b) For y₁ = x,

(43)

For y₂ = x²,

(44)

For

(45)

we have

(46)

Therefore

J[y3]	= 36 ∫ 01x2(1 − x)2 dx	(47)
	= 36 ∫ 01(x2 − 2x3 + x4) dx	(48)
	= 36	(49)
	= 36 = .	(50)

Hence

(51)

(c) Among these three trial curves, the line y₁ = x has the smallest value.

(d) No. This proves only that y₁ beats the two specific competitors chosen here. The full admissible class contains infinitely many other curves. A global-minimum proof must compare against every admissible function or use a theorem that guarantees the result.

CV01 itself gave such a direct proof by writing an arbitrary admissible curve as y = x + u with u(0) = u(1) = 0, which yields

(52)

Solution 5: arc length of a line and a parabola

(a) For y₁ = x, y₁′ = 1, so

(53)

(b) For y₂ = x², y ₂′ = 2x, hence

(54)

Use the standard antiderivative

(55)

Evaluating from 0 to 1 gives

(56)

Since

(57)

we obtain

(58)

(c) Numerically,

(59)

The straight segment is shorter. This agrees with Euclidean geometry: among all sufficiently regular curves joining two fixed points in the plane, the straight line has the least length.

Solution 6: restrict the problem to a one-parameter family

The family is

(60)

Differentiate:

(61)

Then

J(c)	= ∫ 01 2dx	(62)
	= ∫ 01dx.	(63)

The linear term integrates to zero:

(64)

and

(65)

Therefore

(66)

(b) Ordinary calculus gives

(67)

Thus the unique minimum within this family occurs at

(68)

(c) The corresponding curve is

(69)

(d) We have proved that y = x minimizes J among the restricted one-parameter family y_c. By itself, this does not prove that it minimizes over the entire infinite-dimensional admissible class. Restricted trial families are useful approximations, but the conclusion is only as broad as the family searched.

This finite-dimensional reduction foreshadows the Ritz method developed much later in the course.

Solution 7: a quadratic energy functional

For

(70)

we have

(71)

Substitute into the energy:

E(A)	= ∫ 01dx	(72)
	= A2.	(73)

Hence

(74)

Therefore

(75)

(76)

and

(77)

The factor-of-four scaling follows because both terms in the functional are quadratic in y or its derivative. Linear elasticity, small oscillations, and many linear field theories produce quadratic energies for exactly this reason.

Solution 8: evaluating a mechanical action on trial paths

The trial path is

(78)

so

(79)

Substitute into the action:

(80)

The needed integrals are

(81)

and

(82)

Therefore

S(a)	= −	(83)
	= (10m − k) .	(84)

(b) For every nonzero a, the sign is controlled by 10m − k:

(85)

while

(86)

(c) If k = 10m, then

(87)

for every amplitude a in this particular trial family.

(d) Mechanical action contains kinetic energy minus potential energy, not a manifestly positive sum. Stationary-action principles therefore need not be ordinary global minimization statements. Later entries will distinguish stationarity from minimum, maximum, and saddle behavior rigorously.

Solution 9: an infimum that is not attained

For

(88)

we have

(89)

Because c > 0 can be chosen arbitrarily small,

(90)

However, no admissible c > 0 gives J[y_c] = 0. The only parameter that would attain zero is c = 0, and it is excluded from the admissible set. Thus the infimum exists but is not attained.

If the admissible set is changed to

(91)

then c = 0 becomes admissible and the minimum is attained by y = 0.

This simple finite-dimensional example already captures an issue that becomes central in the direct method of the calculus of variations: an infimum and an actual minimizer are different mathematical statements.

3 Summary

These exercises reinforce the basic viewpoint of CV01. A functional is a map from functions to scalars, and its optimization problem is defined not merely by its formula but also by its admissible class. Direct substitution can rank trial curves, sometimes prove a global minimum from positivity, and reveal important physical structure before any variational derivative is introduced.

The restricted-family exercises also anticipate a recurring strategy:

(92)

In CV02 that scalarization is performed infinitesimally with y_𝜖 = y + 𝜖η. In later numerical work it becomes the foundation of Ritz and finite-dimensional approximation methods.