Calculus of Variations: First Variations of Common Functionals

CV02E1 computed first variations directly by expanding J[y + 𝜖η]. This companion set develops the complementary skill: start from the general first-variation formula

∫ b δJ [y;η ] = (F η + F ′η′) dx a y y

(1)

and apply it efficiently to functionals that recur in mechanics, geometry, boundary-value problems, and later variational theory.

The emphasis here is on identifying the correct integrand derivatives and on keeping the first variation separate from the later integration-by-parts step. Where useful, some solutions perform that second step to verify stationarity of a known candidate, but the primary target remains the first variation itself.

Figure. Anatomy of the first variation for a first-order integral functional. Changes in the function itself contribute through F_yη, while changes in slope contribute through F_y′η′.

1 Exercises

Exercise 1: weighted quadratic gradient plus potential

Let

∫ [ ] b 1- ′2 J[y] = 2 a(x)y + V (x,y) dx, a

(2)

where a and V are smooth. Derive δJ[y; η] for a fixed-endpoint variation η.

Exercise 2: forced Dirichlet functional

Consider

∫ 1[1 ] J[y] = --y′2 − f(x )y dx 0 2

(3)

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

Derive the first variation for a general forcing function f(x).
For f(x) = 1, verify that
$y (x) = 1x (1 − x ) ∗ 2$ (4)

is stationary for every fixed-endpoint variation.

Exercise 3: weighted arc-length functional

Let

∫ b ∘ ------- J[y] = w (x,y) 1 + y′2dx, a

(5)

where w(x,y) > 0 is smooth. Compute δJ[y; η]. Then specialize the result to w = 1.

Exercise 4: mechanical action

For a particle with coordinate q(t), define the action

∫ t1[ ] S[q] = 1-m ˙q2 − V(q,t) dt. t0 2

(6)

Derive δS[q; η].
For the harmonic oscillator V (q) = kq², show that any solution of
$m q¨+ kq = 0$ (7)

makes the first variation vanish for all variations satisfying η(t₀) = η(t₁) = 0.

Exercise 5: two coupled dependent variables

Let

∫ b[ 1 1 ] J[u,v] = -u′2 + -v′2 + cuv dx. a 2 2

(8)

Using independent variations η and ξ, compute the full first variation of J[u + 𝜖η,v + 𝜖ξ].

Exercise 6: the p-energy

For p > 1, consider

∫ 1 b ′ p Jp[y ] =-- |y | dx. p a

(9)

Assuming sufficient regularity, derive

δJp [y;η ].

(10)

Check that the formula reduces to the standard quadratic-gradient result when p = 2 and to the quartic result when p = 4.

Exercise 7: adding a total derivative

Let

∫ [ ] ^ b 1-′2 d-- J [y] = 2y + dxG (x,y ) dx, a

(11)

where G is smooth. Show that for fixed-endpoint variations,

∫ b δJ^[y;η] = y′η′dx. a

(12)

Explain why adding a total derivative does not change the interior stationarity condition for fixed endpoint problems.

Exercise 8: explicit x dependence

Consider

∫ [ ] 2 1- 2 ′2 1-x 2 J [y] = 2x y + 2e y dx. 1

(13)

Derive the first variation. For the particular candidate y = 0, determine whether it is stationary for arbitrary fixed-endpoint variations.

2 Solutions

Solution 1: weighted quadratic gradient plus potential

The integrand is

1 F (x,y,y′) = -a(x )y ′2 + V(x, y). 2

(14)

Its relevant partial derivatives are

Fy = Vy(x,y), Fy′ = a (x)y′.

(15)

Therefore

|--------------------------------------| | ∫ b | |δJ[y;η] = [Vy(x, y)η + a(x)y′η ′]dx. | ------------a--------------------------

(16)

This single formula includes many standard one-dimensional energy functionals. The first term measures sensitivity to displacement of the curve; the second measures sensitivity to its slope.

Solution 2: forced Dirichlet functional

Here

F (x, y,y′) = 1y′2 − f (x)y. 2

(17)

Thus

′ ′ Fy = − f (x ), Fy = y .

(18)

The first variation is

|------------------------------| | ∫ 1 | |δJ[y;η] = [y′η′ − f(x )η ]dx. ------------0-------------------

(19)

For f = 1 and

y (x ) = 1x(1 − x), ∗ 2

(20)

one has

′ 1- ′′ y∗ = 2 − x, y∗ = − 1.

(21)

Then

∫ 1[ ( 1 ) ] δJ[y∗;η] = -− x η′ − η dx. 0 2

(22)

Integrating the first term by parts,

∫ ∫ 1 ′ ′ ′ 1 1 ′′ y∗ηdx = [y∗η]0 − y∗ηdx. 0 0

(23)

The boundary term vanishes because η(0) = η(1) = 0, and y_∗′′ = −1, so

∫ ∫ 1 ′ ′ 1 y∗η dx = ηdx. 0 0

(24)

Therefore

|------------| -δJ[y∗;η] =-0-

(25)

for every admissible variation.

Solution 3: weighted arc-length functional

The integrand is

′ ∘ -----′2- F (x,y,y ) = w (x, y) 1 + y .

(26)

Differentiate with respect to y:

∘ ------- Fy = wy(x,y ) 1 + y′2.

(27)

Differentiate with respect to y′:

----y′---- Fy ′ = w (x,y)∘ 1 + y′2.

(28)

Hence

|----------∫-b[-----------------------------]----| | ∘ ------′2 ----y′--- ′ | |δJ[y;η] = a wy 1 + y η + w ∘ 1 + y′2η dx. | --------------------------------------------------

(29)

When w = 1, the displacement term disappears and

|----------------------------| | ∫ b y′ | |δJ[y;η] = ∘--------η′dx. | ------------a----1-+-y′2------|

(30)

This is the first-variation starting point for the shortest-path problem.

Solution 4: mechanical action

The mechanical integrand is the Lagrangian

1 2 L (q, ˙q,t) =-m ˙q − V (q,t). 2

(31)

Under

q𝜖 = q + 𝜖η, ˙q𝜖 = ˙q + 𝜖˙η,

(32)

the first variation is

|----------------------------------| | ∫ t1 | |δS[q;η] = [m ˙q ˙η − Vq(q,t)η]dt. ------------t0----------------------

(33)

For the harmonic oscillator,

Vq = kq,

(34)

∫ t1 δS = [m ˙q˙η − kqη]dt. t0

(35)

Integrate the first term by parts:

∫ ∫ t1 t1 t1 m ˙q˙ηdt = [m ˙qη]t0 − m ¨qηdt. t0 t0

(36)

Because the endpoint variations vanish,

∫ t1 δS = − (m q¨+ kq )ηdt. t0

(37)

Any trajectory satisfying

m q¨+ kq = 0

(38)

therefore gives

|-------| δS-=--0--

(39)

for every admissible η.

Figure. Hamilton’s principle uses exactly the same first-variation machinery as the abstract functional problem: perturb the trajectory while fixing the endpoint events, then differentiate the action with respect to the perturbation amplitude.

Solution 5: two coupled dependent variables

Perturb both variables:

u = u + 𝜖η, v = v + 𝜖ξ. 𝜖 𝜖

(40)

Then

′ ′ ′ ′ ′ ′ u𝜖 = u + 𝜖η , v𝜖 = v + 𝜖ξ .

(41)

Differentiating the scalarized functional at 𝜖 = 0 gives

|---------------∫------------------------------| | b ′ ′ ′ ′ | |δJ [u,v; η,ξ] = [uη + v ξ + c(vη + uξ)]dx. | -----------------a-----------------------------

(42)

The two variations are independent. Later, the vector version of the Euler–Lagrange equation will turn the coefficients of η and ξ into a coupled system of differential equations.

Solution 6: the p-energy

Let

′ 1- ′p F (y ) = p|y |.

(43)

For p > 1, the derivative with respect to y′ is

′p−2 ′ Fy′ = |y | y .

(44)

There is no explicit dependence on y, so F_y = 0. Therefore

|----------∫-b--------------| | ′ p− 2 ′′ | δJp[y;η] = a |y | y η dx. | -----------------------------

(45)

For p = 2,

∫ b ′′ δJ2[y;η] = yη dx. a

(46)

For p = 4,

∫ b δJ4[y;η] = y′3η′dx. a

(47)

If instead one omits the prefactor 1∕p and uses J = ∫ y′⁴dx, the corresponding first variation is 4 ∫ y′³η′dx, matching the direct expansion in CV02E1.

Solution 7: adding a total derivative

Write

∫ b1 ^J[y] = -y ′2dx + G (b,y(b)) − G (a,y(a)). a 2

(48)

Under a variation, the first term contributes

∫ b ′ ′ y η dx. a

(49)

The boundary contribution varies as

δ[G(b,y(b)) − G (a, y(a))] = Gy (b,y(b))η (b) − Gy (a,y(a))η(a).

(50)

For fixed endpoints,

η(a) = η(b) = 0,

(51)

so this term vanishes. Hence

|--------------------| | ∫ b | |δ ^J[y;η] = y′η′dx.| ------------a---------

(52)

Thus a total derivative changes only endpoint data. When endpoint values are fixed, it does not alter the interior stationarity equation. In mechanics, this is the familiar fact that Lagrangians differing by a total time derivative produce the same equations of motion.

Figure. Adding a total derivative changes the action by a boundary contribution. For fixed endpoints its first variation vanishes, leaving the interior variational condition unchanged.

Solution 8: explicit x dependence

The integrand is

F(x, y,y′) = 1x2y′2 + 1exy2. 2 2

(53)

Therefore

x ′ 2 ′ Fy = e y, Fy = x y.

(54)

The first variation is

|--------------------------------| | ∫ 2[ ] | |δJ[y;η] = exy η + x2y′η ′dx. | ------------1---------------------

(55)

For y = 0, both y and y′ vanish identically, so

|------------| |δJ [0;η] = 0 | -------------

(56)

for every admissible variation. Thus y = 0 is stationary. Because the integrand is nonnegative and vanishes at y = 0, one can in this special case go further and identify it as a global minimizer, but that conclusion uses more than the first variation alone.

3 Summary

For the classical first-order functional

∫ b J [y] = F (x,y,y ′)dx, a

(57)

the practical workflow is

∫ b F −→ (Fy,Fy ′) −→ δJ [y;η] = (Fyη + Fy ′η ′)dx. a

(58)

The examples show that the same structure covers elastic-energy-like functionals, forced boundary-value problems, path length, mechanical action, coupled coordinates, nonlinear p-energies, and Lagrangians modified by total derivatives. The first variation is therefore the common language connecting a large class of physical extremum principles.

"Calculus of Variations: First Variations of Common Functionals" is owned by bloftin.
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