Calculus of Variations Examples: Fundamental Lemma Problems

This companion entry develops the Fundamental Lemma through calculation rather than another theorem presentation. The exercises move from hypothesis checks to explicit counterexamples, then to scalar and vector Euler–Lagrange-type integral identities and finally to the weak almost-everywhere form.

The convention throughout is that a statement of the form

(1)

for every admissible test function carries much more information than the same equality for one or finitely many chosen functions.

1 Exercises

Exercise 1: Hypothesis audit

For each statement below, determine the strongest conclusion that follows and state which version of the Fundamental Lemma is being used.

1. g ∈ C([0, 1]) and

   (2)

   for every η ∈ C_c^∞(0, 1).

2. g ∈ C([0, 1]) and

   (3)

3. g ∈ C([0, 1]) and

   (4)

   for every η ∈ C¹([0, 1]) satisfying η(0) = η(1) = 0.

4. g ∈ L_loc¹(0, 1) and

   (5)

   for every η ∈ C_c^∞(0, 1).

Exercise 2: One test function is not enough

Let

(6)

Find a nonzero linear function

(7)

such that

(8)

Explain why this does not contradict the Fundamental Lemma.

Exercise 3: A weighted-square proof

Suppose g ∈ C¹([0, 1]) and

(9)

for every η ∈ C¹([0, 1]) satisfying η(0) = η(1) = 0. Without invoking the Fundamental Lemma by name, choose a test function depending on g and prove directly that g ≡ 0.

Exercise 4: Recover a differential equation and solve it

Suppose y ∈ C²([0, 1]) satisfies

(10)

for every η ∈ C_c^∞(0, 1), together with

(11)

Use the Fundamental Lemma to determine y(x).

Exercise 5: First variation to boundary-value problem

Let y ∈ C²([0, 1]) satisfy the fixed-endpoint stationarity identity

(12)

for every η ∈ C¹([0, 1]) with η(0) = η(1) = 0. If

(13)

derive and solve the differential equation for y.

Exercise 6: Vector-valued Fundamental Lemma

Let u,v ∈ C¹([0, 1]) and suppose

(14)

for every pair (η₁,η₂) of smooth compactly supported test functions.

1. Derive the coupled differential equations for u and v.
2. Solve them subject to

   (15)

Exercise 7: Pointwise versus almost-everywhere equality

Define

(16)

Show that

(17)

for every smooth compactly supported test function η. Why is the conclusion g = 0 almost everywhere correct while the conclusion g = 0 at every point is false?

Exercise 8: Stationarity is not classification

Consider

(18)

on the fixed-endpoint class y(0) = y(1) = 0.

1. Show that y_∗(x) = 0 is stationary.
2. Classify y_∗ globally.
3. Explain which part of the reasoning uses the Fundamental Lemma and which part does not.

2 Solutions

Solution 1: Hypothesis audit

(a) The classical Fundamental Lemma applies directly. Since g is continuous and the integral vanishes for every smooth compactly supported test function,

(19)

for every x ∈ (0, 1), and continuity extends the conclusion to the endpoints. Thus

(20)

(b) No such conclusion follows. This is one scalar orthogonality condition on an infinite-dimensional set of possible g. A nonzero function can easily have zero weighted integral against a single chosen test function. Exercise 2 constructs one explicitly.

(c) The fixed-endpoint corollary applies. Every function in C_c^∞(0, 1) is also a C¹ function that vanishes at both endpoints, so the hypothesis includes the test class required by the Fundamental Lemma. Therefore

(21)

(d) The weak Fundamental Lemma applies. The correct conclusion is

(22)

for almost every x ∈ (0, 1). Without additional regularity such as continuity, pointwise equality everywhere cannot be inferred.

Solution 2: One test function is not enough

We require

(23)

Separate the two terms:

(24)

The required integrals are

(25)

and

(26)

Therefore

(27)

so

(28)

Thus

(29)

is nonzero yet satisfies the given integral identity.

There is no contradiction with the Fundamental Lemma because the lemma assumes that the integral vanishes for every test function in a sufficiently rich class, not for one chosen η.

Solution 3: A weighted-square proof

Choose

(30)

Because g ∈ C¹([0, 1]), this η belongs to C¹([0, 1]), and

(31)

It is therefore an admissible test function. Substitution gives

(32)

For 0 < x < 1,

(33)

and g(x)² ≥ 0. Hence the integrand is nonnegative. A continuous nonnegative function with zero integral must vanish identically, so

(34)

for every x ∈ [0, 1]. On the open interval this implies

(35)

Continuity then gives g(0) = g(1) = 0. Therefore

(36)

This proof is efficient, but it relies on the test class being broad enough to allow a test function constructed from g itself.

Solution 4: Recover a differential equation and solve it

Define

(37)

Since y ∈ C², the function q is continuous. The hypothesis says

(38)

for every smooth compactly supported η. The Fundamental Lemma therefore gives

(39)

or

(40)

Integrating once,

(41)

Integrating again,

(42)

The condition y(0) = 0 gives C₂ = 0. The condition y(1) = 1 gives

(43)

hence

(44)

Therefore

(45)

Solution 5: First variation to boundary-value problem

Begin with

(46)

Integrate the second term by parts:

(47)

Since η(0) = η(1) = 0, the boundary term vanishes. Thus

(48)

for every fixed-endpoint test function. The Fundamental Lemma gives

(49)

or

(50)

The general solution is

(51)

From y(0) = 0,

(52)

From y(1) = 1,

(53)

Hence

(54)

This exercise exhibits the complete chain that CV03 supports:

(55)

Solution 6: Vector-valued Fundamental Lemma

Because the test components are independently arbitrary, choose first

(56)

Then

(57)

for every scalar test function η₁. Hence

(58)

Now choose

(59)

Then

(60)

so

(61)

Thus the coupled system is

(62)

Differentiate u′ = v to obtain

(63)

so

(64)

Hence

(65)

Since v = u′,

(66)

The initial data u(0) = 1 and v(0) = 0 give

(67)

Therefore

(68)

Solution 7: Pointwise versus almost-everywhere equality

The function g differs from zero at only the single point x = 1∕2. Changing an integrand at a set of measure zero does not change its integral. Therefore for every test function η,

(69)

Nevertheless,

(70)

Thus pointwise equality everywhere is false. However, the set on which g is nonzero has measure zero, so

(71)

This is exactly why the weak Fundamental Lemma concludes almost-everywhere equality for locally integrable functions. Continuity is what upgrades that conclusion to pointwise equality in the classical theorem.

Solution 8: Stationarity is not classification

For

(72)

the first variation is

(73)

At y_∗ = 0,

(74)

so

(75)

for every admissible η. Thus y_∗ = 0 is stationary.

For every admissible y,

(76)

At y = 0,

(77)

Therefore

(78)

The Fundamental Lemma is used when converting a stationary integral identity into a pointwise differential equation. It does not classify a stationary function as a minimum or maximum. Classification here comes from the sign of the functional itself.

3 Summary

The exercises reinforce four distinct lessons:

1. the Fundamental Lemma requires a rich class of test functions;
2. one or finitely many vanishing weighted integrals do not imply pointwise vanishing;
3. scalar and vector integral identities become pointwise differential equations through localization; and
4. the weak theorem naturally gives almost-everywhere equality, while classification of stationary curves requires additional arguments.