Calculus of Variations: Computing Second Variations - Exercises

Calculus of Variations: Computing Second Variations - Exercises and Worked Solutions

CV11 introduced the second variation as the variational analogue of a Hessian quadratic form. This companion entry is deliberately computational. The goal is to become fluent at moving between three equivalent viewpoints:

(1)

(2)

and, for a first-order functional J[y] = ∫ _a^bF(x,y,y′)dx,

(3)

All derivatives of F are evaluated along the candidate y_∗. The main lesson emphasized that this quadratic form carries classification information only after y_∗ has first been shown to be stationary. The exercises below reinforce that separation. Standard treatments of the second variation and Legendre condition include [1, 2, 3, 4].

1 Exercises

Exercise 1: a positive quadratic functional

Consider

(4)

where ω is a real constant.

1. Verify that y_∗(x) = 0 is stationary.
2. Compute δ²J[0; η] from the Hessian formula.
3. Evaluate the result for η(x) = sin(πx).

Exercise 2: direct epsilon expansion with variable coefficients

Let

(5)

For a general candidate y_∗ and variation η:

1. substitute y_∗ + 𝜖η directly and isolate the coefficient of 𝜖²;
2. infer δ²J[y _∗; η];
3. check the result using F_yy, F_yy′, and F_y′y′;
4. evaluate it for η = x(1 − x).

Exercise 3: a mixed term that becomes a boundary term

Consider

(6)

with fixed endpoints y(0) = y(1) = 0.

1. Compute the three Hessian coefficients.
2. Write the unreduced second variation.
3. Show explicitly that the mixed term integrates to zero for admissible variations.
4. Evaluate the reduced form for η = sin(πx).

Exercise 4: a nontrivial stationary logarithmic extremal

Consider

(7)

with endpoint values

(8)

1. Verify that

   (9)

   is an Euler–Lagrange extremal.

2. Compute F_yy, F_yy′, and F_y′y′ along y_∗.
3. Write the unreduced second variation.
4. Integrate the mixed term by parts and show that the η² term cancels completely.

Exercise 5: second variation of arc length about a straight line

For

(10)

let

(11)

join the prescribed endpoints.

1. Compute the second variation about y_∗.
2. Evaluate it for

   (12)

3. State what the sign tells you and what it does not prove by itself.

Exercise 6: harmonic-oscillator action and trial modes

For the mechanical action

(13)

let q_∗ be any stationary trajectory with fixed endpoints.

1. Compute δ²S[q _∗; η].
2. Evaluate it on the trial family

   (14)

3. For each n, find the interval length at which this particular trial direction becomes neutral.

Exercise 7: a nonlinear slope energy and the Legendre coefficient

Let

(15)

1. Compute the second variation about a stationary function y_∗.
2. Identify the Legendre coefficient F_y′y′.
3. If y_∗ = 0 is stationary and V ′(0) = 0, simplify the result.
4. Explain immediately why α < 0 rules out y_∗ = 0 as a weak local minimum.

Exercise 8: a variable-coefficient quadratic form

Consider

(16)

1. Show that y_∗ = 0 is stationary and compute its second variation.
2. Evaluate the second variation on η = sin(πx).
3. Find the values of λ for which this one trial direction already proves that y_∗ = 0 cannot be a local minimum.

Exercise 9: the bilinear variational Hessian

For

(17)

define

(18)

1. Compute B(η,ξ).
2. Verify explicitly that B(η,ξ) = B(ξ,η).
3. Evaluate B for

   (19)

4. Explain the relation between B(η,η) and the ordinary second variation.

2 Worked solutions

Solution 1: a positive quadratic functional

The integrand is

(20)

Its first derivatives are

(21)

At y_∗ = 0, both vanish, so the Euler–Lagrange equation is satisfied. The Hessian coefficients are

(22)

Therefore

(23)

For η = sin(πx),

(24)

and

(25)

Hence

(26)

This is strictly positive for every real ω and every nonzero version of this trial direction.

Solution 2: direct epsilon expansion with variable coefficients

Substitute

(27)

Then

J[y𝜖]	= ∫ 01dx	(28)
	= J[y∗] + 𝜖∫ 01 dx	(29)
	+ 𝜖2 ∫ 01dx.	(30)

Because the Taylor convention is

(31)

we obtain

(32)

The Hessian check gives

(33)

which reproduces the same result.

For

(34)

we find

δ2J	= ∫ 01dx	(35)
	= .	(36)

This example is useful because the second variation does not depend on y_∗ at all: the original functional is already quadratic in y and y′.

Solution 3: a mixed term that becomes a boundary term

For

(37)

we have

(38)

Therefore

(39)

Now use

(40)

Thus

(41)

Because admissible fixed-endpoint variations satisfy

(42)

this boundary contribution vanishes. Hence

(43)

For η = sin(πx),

(44)

Notice that α disappears completely from the fixed-endpoint second variation. In fact, αyy′ = (d∕dx)(y²) is itself a total derivative.

Solution 4: a nontrivial stationary logarithmic extremal

Here

(45)

The first derivatives are

(46)

Therefore

(47)

The Euler–Lagrange equation becomes

(48)

For

(49)

we have

(50)

so indeed

(51)

The second derivatives of F are

(52)

(53)

and

(54)

Along y_∗ = ln(1 + x),

(55)

and therefore

(56)

The unreduced form is

(57)

Since

(58)

integration by parts gives

(59)

The boundary term vanishes, so the −2η² produced here cancels the original +2η². Thus

(60)

This is a useful example because a complicated-looking Hessian quadratic form collapses to a manifestly nonnegative expression after the mixed term is handled correctly.

Solution 5: second variation of arc length about a straight line

For

(61)

we have

(62)

Also,

(63)

so

(64)

Along the straight line y_∗′ = m,

(65)

For

(66)

we have

(67)

and therefore

(68)

Hence

(69)

This positive value confirms positive curvature of the functional along this trial direction. Positivity in one direction alone does not prove a minimum. Here the full quadratic form is positive for every nonzero fixed-endpoint variation, which is much stronger; a complete sufficiency statement still belongs to the later theory.

Solution 6: harmonic-oscillator action and trial modes

For

(70)

we have

(71)

Thus every stationary trajectory has the same second variation,

(72)

Take

(73)

Then

(74)

and

(75)

Therefore

(76)

This direction becomes neutral when

(77)

or

(78)

For T > T_n, that trial direction produces a negative second variation. The first mode n = 1 therefore detects loss of positive definiteness first. This is the calculation behind the conjugate-point threshold developed more systematically in CV12 and is standard in variational mechanics [5, 1].

Solution 7: a nonlinear slope energy and the Legendre coefficient

The integrand is

(79)

Its relevant second derivatives are

(80)

and

(81)

Therefore about a stationary y_∗,

(82)

The Legendre coefficient is

(83)

If y_∗ = 0 is stationary, then V ′(0) = 0 and y_∗′ = 0, so

(84)

If α < 0, then

(85)

along the candidate. The Legendre necessary condition for a weak local minimum is violated immediately. No further classification calculation is needed to rule out a weak minimum.

Solution 8: a variable-coefficient quadratic form

The functional is already quadratic, and y_∗ = 0 plainly satisfies the Euler–Lagrange equation. The Hessian coefficients are

(86)

Hence

(87)

For η = sin(πx),

(88)

Also,

(89)

so

∫ 01(1 + x)(η′)2dx	= π2 ∫ 01(1 + x) cos 2(πx)dx	(90)
	= .	(91)

Therefore

(92)

This trial direction is negative when

(93)

Thus

(94)

This is only a one-direction test. If the value is positive, we have not proved that the quadratic form is positive for every admissible variation.

Solution 9: the bilinear variational Hessian

For the quadratic functional,

(95)

The integrand is unchanged when η and ξ are interchanged, so

(96)

Now choose

(97)

Orthogonality gives

(98)

Their derivatives are proportional to cos(πx) and cos(2πx), which are also orthogonal on [0, 1], so

(99)

Hence

(100)

Finally, the ordinary quadratic second variation is simply the bilinear Hessian evaluated twice in the same direction:

(101)

This is exactly analogous to the finite-dimensional expression h^T Hh obtained by inserting the same displacement vector into both slots of the Hessian bilinear form.

3 A compact computation checklist

For a functional

(102)

a reliable calculation proceeds as follows.

1. Verify that the candidate is stationary before using the second variation for classification.
2. Compute

   (103)

   along the candidate.

3. Assemble

   (104)

4. If useful, reduce the mixed term using

   (105)

5. Apply the actual endpoint conditions before discarding boundary terms.
6. Check the Legendre coefficient C = F_y′y′.
7. Evaluate selected trial functions to find negative or neutral directions.
8. Do not infer positivity for all variations from positivity on only one or a few trial directions.

The last point is the bridge to CV11E2 and CV12: computing a quadratic form is straightforward; proving its sign for an infinite-dimensional class is the deeper problem.

4 Summary

The second variation can be obtained either by direct epsilon expansion or by differentiating the integrand Hessian. For fixed-endpoint problems the general expression is

(106)

The mixed term often simplifies after integration by parts, but it must not simply be dropped. The Legendre coefficient F_y′y′ gives a pointwise necessary condition, while selected trial directions can disprove minimality by producing a negative second variation. Positive values on a few directions are not a proof of a minimum. That distinction motivates the classification and Jacobi-field problems that follow.

References