Calculus of Variations: Direct Epsilon Expansion Exercises

Calculus of Variations: Direct 𝜖-Expansion Exercises

This companion to CV02 develops fluency with the definition

d || δJ [y;η ] = --J[y + 𝜖η]|| d𝜖 𝜖=0

(1)

without beginning from the memorized first-variation formula. The purpose is to make the scalarization step automatic: construct y_𝜖, substitute it into the functional, expand in powers of 𝜖, and read the first-order coefficient.

For a sufficiently smooth scalarization,

𝜖2 2 Φ (𝜖) = J[y + 𝜖η] = J [y] + 𝜖δJ [y;η] + 2 δ J[y;η] + ⋅⋅⋅ .

(2)

Thus the coefficient of 𝜖 is the first variation in the chosen direction. These exercises also show why a vanishing linear term establishes stationarity in a direction but does not, by itself, classify the candidate.

Figure. The scalar expansion of a functional along one variation direction. The constant term is the original functional value, the linear coefficient is the first variation, and the quadratic coefficient contains second-order information.

1 Exercises

Exercise 1: endpoint admissibility

Let

y𝜖(x ) = y(x) + 𝜖η (x )

(3)

on [a,b], where the admissible curves satisfy fixed endpoint conditions y(a) = A and y(b) = B.

Show that y_𝜖 satisfies the same endpoint conditions for every sufficiently small 𝜖 if η(a) = η(b) = 0.
Show conversely that if y_𝜖(a) = A and y_𝜖(b) = B for all 𝜖 in an interval containing zero, then necessarily η(a) = η(b) = 0.

Exercise 2: stationary straight line by direct expansion

Consider

∫ 1 ′ 2 J [y] = y(x) dx 0

(4)

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

y(x) = x, η(x) = x(1 − x).

(5)

Construct Φ(𝜖) = J[y + 𝜖η] exactly. Find Φ(0), Φ′(0), and Φ′′(0). Interpret the result.

Exercise 3: detecting a nonstationary curve

Use the same functional and endpoint conditions as Exercise 2, but now take

y(x) = x2, η(x) = x (1 − x ).

(6)

Compute Φ(𝜖) exactly and determine δJ[y; η]. Explain why one nonzero first variation is sufficient to rule out stationarity.

Exercise 4: a functional depending only on y

Let

∫ 1 J [y] = y(x)2dx, 0

(7)

and choose

y(x) = x, η(x) = x(1 − x).

(8)

Compute the exact polynomial Φ(𝜖) = J[y + 𝜖η] and find the first variation.

Exercise 5: first-order stationarity with positive quadratic change

For

∫ 1( ′2 2) J[y] = y + y dx, 0

(9)

use the candidate y = 0 and the fixed-endpoint variation

η(x) = sin(πx ).

(10)

Find Φ(𝜖) exactly. What do the first and second derivatives at zero say?

Exercise 6: nonlinear slope functional

Let

∫ 1 ′ 4 J[y] = y (x) dx, 0

(11)

with y(x) = x and η(x) = x(1 − x). Expand J[y + 𝜖η] through all powers of 𝜖 and compute δJ[y; η].

Exercise 7: a total-derivative functional

Let

∫ 1 ′ J[y] = y(x)y (x)dx 0

(12)

on the fixed-endpoint class y(0) = 0, y(1) = 1. For an arbitrary admissible variation η, directly expand

J [y + 𝜖η]

(13)

and prove that the entire scalarized functional is independent of 𝜖. Explain the result using the identity yy′ = d-
dx (y²∕2).

Exercise 8: two directions and linearity

Consider

∫ 1 ( 2 ′2) J [y] = y + y dx 0

(14)

at y(x) = x, with fixed-endpoint directions

η1(x) = x(1 − x), η2(x ) = sin(πx ).

(15)

Compute δJ[x; η₁] and δJ[x; η₂] directly.
Let η = αη₁ + βη₂. Use a direct 𝜖-expansion to verify
$δJ [x; η] = αδJ [x;η1] + βδJ [x; η2].$ (16)

2 Solutions

Solution 1: endpoint admissibility

At the left endpoint,

y 𝜖(a) = y(a) + 𝜖η(a).

(17)

If y(a) = A and η(a) = 0, then y_𝜖(a) = A for every 𝜖. The same argument gives y_𝜖(b) = B when η(b) = 0.

Conversely, assume y_𝜖(a) = A for every 𝜖 near zero. Since the unperturbed curve is admissible,

A = y𝜖(a) = A + 𝜖η(a).

(18)

Hence

𝜖η(a) = 0

(19)

for nonzero 𝜖 as well as zero, so η(a) = 0. Identically, η(b) = 0.

Thus fixed endpoint constraints translate directly into homogeneous endpoint conditions on admissible variations.

Solution 2: stationary straight line by direct expansion

The perturbed curve is

y𝜖 = x + 𝜖x(1 − x),

(20)

and

y ′𝜖 = 1 + 𝜖(1 − 2x).

(21)

Therefore

∫ 1 2 Φ (𝜖) = [1 + 𝜖(1 − 2x)] dx. 0

(22)

Expanding,

∫ 1 ∫ 1 Φ (𝜖) = 1 + 2𝜖 (1 − 2x )dx + 𝜖2 (1 − 2x)2dx. 0 0

(23)

The linear integral is zero, while

∫ 1 ∫ 1 1 (1 − 2x )2dx = (1 − 4x + 4x2)dx = -. 0 0 3

(24)

Thus

|--------------| | 𝜖2 | |Φ (𝜖) = 1 + 3 | ---------------

(25)

and

Φ (0 ) = 1, Φ ′(0) = 0, Φ′′(0) = 2-. 3

(26)

The zero first derivative says that the straight line is stationary in this direction. The positive second derivative says that along this particular one-parameter family, 𝜖 = 0 is a strict local minimum. This does not yet prove a minimum against every admissible variation, although for this functional the straight line is in fact the global minimizer.

Solution 3: detecting a nonstationary curve

Now

y = x2 + 𝜖x(1 − x), 𝜖

(27)

y′𝜖 = 2x + 𝜖(1 − 2x).

(28)

Hence

∫ 1 2 Φ (𝜖) = [2x + 𝜖(1 − 2x)] dx. 0

(29)

Expanding,

∫ 1 ∫ 1 ∫ 1 Φ (𝜖) = 4x2dx + 4𝜖 x (1 − 2x )dx + 𝜖2 (1 − 2x)2dx. 0 0 0

(30)

The three pieces are

∫ 1 2 4- 0 4x dx = 3,

(31)

∫ 1 ( 1 2 ) 2 4 x(1 − 2x)dx = 4 --− -- = − --, 0 2 3 3

(32)

and the final integral is 1∕3. Therefore

|-------4---2-----1--| Φ (𝜖) = --− --𝜖 + -𝜖2| --------3---3-----3---

(33)

and

|------------------------| | 2 ′ 2- | |δJ [x ;η ] = Φ (0) = − 3 . -------------------------

(34)

Stationarity requires the first variation to vanish for every admissible direction. Finding even one admissible η with a nonzero first variation therefore disproves stationarity immediately.

Figure. Two scalarized functionals. The straight-line candidate has a horizontal tangent at 𝜖 = 0, whereas the x² candidate has nonzero slope there. The latter is therefore not stationary.

Solution 4: a functional depending only on y

Substitute

y = x + 𝜖x(1 − x). 𝜖

(35)

Then

∫ 1 2 Φ(𝜖) = [x + 𝜖x (1 − x )] dx. 0

(36)

Expand:

∫ 1 ∫ 1 ∫ 1 Φ (𝜖) = x2dx + 2𝜖 x2(1 − x)dx + 𝜖2 x2(1 − x)2dx. 0 0 0

(37)

The integrals are

∫ 1 2 1- x dx = 3 , 0

(38)

∫ 1 ( 1 1 ) 1 2 x2(1 − x)dx = 2 --− -- = --, 0 3 4 6

(39)

and

∫ 1 2 2 1- 1- 1- -1- x (1 − x )dx = 3 − 2 + 5 = 30 . 0

(40)

Therefore

|----------------------| |Φ (𝜖) = 1-+ 1𝜖 + -1𝜖2 | --------3----6----30---|

(41)

and

|------------| | 1-| δJ [x;η] = 6.| --------------

(42)

Solution 5: first-order stationarity with positive quadratic change

With y = 0 and η = sin(πx),

y𝜖 = 𝜖 sin(πx ), y′𝜖 = 𝜖π cos(πx).

(43)

Thus

∫ 2 1 [ 2 2 2 ] Φ(𝜖) = 𝜖 π cos (πx ) + sin (πx) dx. 0

(44)

Both squared trigonometric terms integrate to 1∕2, so

|--------2-------| Φ (𝜖) = π--+-1𝜖2.| ----------2-------

(45)

Consequently,

Φ′(0) = 0, Φ ′′(0) = π2 + 1 > 0.

(46)

The candidate is stationary in this direction and the cost rises quadratically for small positive or negative 𝜖. The first derivative supplies the stationarity information; the second derivative begins the classification question developed later in the series.

Solution 6: nonlinear slope functional

Again let

q(x) = 1 − 2x,

(47)

so that y_𝜖′ = 1 + 𝜖q. Then

∫ 1 Φ(𝜖) = (1 + 𝜖q)4dx. 0

(48)

The binomial expansion gives

(1 + 𝜖q)4 = 1 + 4𝜖q + 6𝜖2q2 + 4𝜖3q3 + 𝜖4q4.

(49)

By symmetry about x = 1∕2,

∫ ∫ 1 1 3 q dx = 0, q dx = 0. 0 0

(50)

Also,

∫ 1 1 ∫ 1 1 q2dx = --, q4dx = --. 0 3 0 5

(51)

Hence

|---------------------| | 2 1- 4 | Φ (𝜖) = 1 + 2𝜖 + 5 𝜖. | -----------------------

(52)

In particular,

|------------| -δJ[x;η] =-0.-

(53)

This example shows that direct 𝜖-expansion is not restricted to quadratic functionals.

Solution 7: a total-derivative functional

Expand directly:

∫ 1 ′ ′ J [y + 𝜖η] = (y + 𝜖η)(y + 𝜖η )dx. 0

(54)

Therefore

∫ 1 ∫ 1 J[y + 𝜖η] = J [y ] + 𝜖 (yη ′ + y′η)dx + 𝜖2 ηη′dx. 0 0

(55)

The linear integrand is a total derivative:

′ ′ d-- yη + yη = dx(y η),

(56)

so fixed-endpoint variations give

∫ 1 (yη′ + y′η)dx = [yη]10 = 0. 0

(57)

Likewise,

∫ 1 ′ 1- 21 ηη dx = 2[η ]0 = 0. 0

(58)

Thus

|----------------| |J[y + 𝜖η] = J[y]| -----------------

(59)

for every admissible variation. In fact,

∫ ( ) 1 d-- 1-2 1- 2 2 1- J[y] = dx 2y dx = 2[y(1) − y (0)] = 2 . 0

(60)

The value depends only on the fixed boundary data, not on the interior path.

Solution 8: two directions and linearity

For

∫ 1 J[y] = (y2 + y′2)dx, 0

(61)

the direct first-order expansion about y = x is

∫ 1 ′ 2 J[x + 𝜖η] = J [x] + 2𝜖 (xη + η )dx + O(𝜖 ). 0

(62)

Because every admissible variation vanishes at the endpoints,

∫ 1 ′ η dx = 0. 0

(63)

For η₁ = x(1 − x),

∫ 1 |--| δJ[x;η1] = 2 x2(1 − x)dx = |1-. 0 -6-|

(64)

For η₂ = sin(πx),

∫ 1 δJ[x;η2] = 2 xsin(πx )dx. 0

(65)

Integration by parts gives

∫ 1 1 x sin(πx )dx = -, 0 π

(66)

|-----------2--| |δJ[x;η2] = --.| ------------π--|

(67)

Now let

η = αη1 + βη2.

(68)

The coefficient of 𝜖 in the direct expansion is

∫ 1 1- 2- 2 x (α η1 + βη2)dx = α 6 + β π. 0

(69)

Therefore

|------------------------------------------| -δJ[x;α-η1 +-β-η2] =-α-δJ-[x;-η1] +-β-δJ[x;η2].

(70)

The linearity of the first variation is therefore visible directly in the coefficient of the scalar perturbation parameter.

3 Summary

The exercises reinforce a reliable direct procedure:

choose an admissible perturbation η;
form y_𝜖 = y + 𝜖η;
substitute y_𝜖 and its derivatives into J;
expand in powers of 𝜖; and
identify the coefficient of 𝜖 as δJ[y; η].

A zero linear coefficient means stationarity in that direction. A nonzero coefficient rules out stationarity immediately. Higher powers of 𝜖 contain classification information, but the first variation itself is strictly a first-order object.

"Calculus of Variations: Direct Epsilon Expansion Exercises" is owned by bloftin.
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