Calculus of Variations: First Integrals and Special Forms of the Euler--Lagrange Equation

Calculus of Variations: First Integrals and Special Forms of the Euler–Lagrange Equation

The Euler–Lagrange equation is usually a second-order differential equation. That does not mean every variational problem must be attacked as a generic second-order boundary-value problem. When the integrand does not depend explicitly on one of its variables, the Euler–Lagrange equation contains a hidden first integral: a quantity that remains constant along every sufficiently smooth stationary curve.

These reductions are among the most useful computational shortcuts in the classical calculus of variations. They also provide the first clear glimpse of a deeper physical principle: invariance of a variational problem is closely connected with a conserved quantity. The full theorem behind that observation is Noether’s theorem, developed later in CV20; here we derive the two elementary cases directly from Euler–Lagrange [1, 2, 3].

Figure. Two special structures produce immediate first integrals. If the dependent variable y is absent explicitly, the conjugate quantity F_y′ is constant. If the independent variable x is absent explicitly, the Beltrami quantity F − y′F_y′ is constant.

1 Learning objectives

After completing CV06, the reader should be able to

recognize when an Euler–Lagrange problem has a cyclic dependent variable;
derive the first integral F_y′ = C when F_y = 0;
derive the du Bois–Reymond identity (F − y′F_y′) = F_x along a stationary curve;
obtain the Beltrami identity when F_x = 0;
understand why a first integral reduces the differential order by one;
use a first integral to convert a variational problem to a quadrature;
connect the autonomous mechanical case to conservation of energy;
distinguish the cases F_y = 0 and F_x = 0; and
recognize these first integrals as elementary precursors of Noether’s theorem.

2 Starting point: the Euler–Lagrange equation

Consider the first-order functional

∫ b ′ J[y] = F (x,y,y )dx, a

(1)

with F sufficiently smooth and with a stationary curve y_∗(x) smooth enough for the derivatives below to exist. CV04 derived the necessary interior condition

|----------------| |F − -d-F ′ = 0.| --y---dx--y-------

(2)

Throughout this article, the partial derivatives of F are evaluated along the candidate curve unless stated otherwise.

A generic equation of this form contains y′′ through

d-F ′ = F ′ + F ′y ′ + F ′ ′y′′. dx y xy yy y y

(3)

If F_y′y′≠0, solving Euler–Lagrange therefore normally requires a second-order differential equation and two boundary conditions. A first integral replaces that second-order equation by a first-order relation.

3 First special form: the dependent variable is cyclic

Suppose the integrand has no explicit dependence on y:

′ F = F (x,y ).

(4)

Then

Fy = 0.

(5)

Euler–Lagrange immediately becomes

-d- − dx Fy′ = 0,

(6)

d-- dxFy ′ = 0.

(7)

Therefore

|--------| -Fy′ =-C.-

(8)

The variable y is called a cyclic or ignorable variable because it does not appear explicitly in F. The derivative

p = Fy ′

(9)

is often called the momentum conjugate to y. In an abstract variational problem this is simply useful terminology; in mechanics it becomes canonical momentum under the usual identification F = L [4, 5].

3.1 Why “cyclic” does not mean constant

A common misconception is that if y is absent from F, then y itself must be constant. That is false. The conserved quantity is

Fy′,

(10)

which may be a nonlinear function of y′ and may also depend explicitly on x.

For example, let

F(x, y′) = 1-(1 + x )(y ′)2. 2

(11)

Then

′ ′ Fy = (1 + x)y .

(12)

The first integral gives

(1 + x )y′ = C,

(13)

′ --C--- y = 1 + x.

(14)

A single integration yields

|------------------------| -y(x)-=-C-ln(1-+-x)-+-D.-|

(15)

The stationary curve need not be constant even though y is cyclic.

4 The du Bois–Reymond identity

A second important reduction begins with the combination

F − y′F ′. y

(16)

Differentiate it along an arbitrary sufficiently smooth curve:

− y′′F_y′− y′ d
---
dx

F_y′.

(17)

Because

F = F (x, y(x),y′(x)),

(18)

the total derivative of F is

dF- = F + F y′ + F ′y′′. dx x y y

(19)

Substituting gives

	= F_x + F_yy′ + F_y′y′′− y′′F_y′− y′F_y′	(20)
	= F_x + y′.	(21)

The two terms containing y′′F_y′ cancel identically. Along a stationary curve, Euler–Lagrange makes the expression in parentheses vanish. Therefore

|--------------------| |d-- ′ | |dx (F − y Fy′) = Fx.| ----------------------

(22)

This relation is commonly called the du Bois–Reymond condition or du Bois–Reymond identity in the classical theory [1, 2]. It is not an additional independent equation when the classical Euler–Lagrange equation already holds with sufficient smoothness; rather, it is a useful consequence of it.

Figure. The derivative of F −y′F_y′ contains two cancellations. The chain-rule terms F_y′y′′ and −y′′F_y′ cancel algebraically; the remaining Euler–Lagrange residual vanishes on a stationary curve.

5 The Beltrami identity

Now suppose the integrand has no explicit dependence on the independent variable:

′ F = F (y,y ).

(23)

Then

Fx = 0.

(24)

The du Bois–Reymond identity reduces to

d ---(F − y′Fy′) = 0. dx

(25)

Hence

|-----′--------| F--−-y-Fy′ =-C.-

(26)

This is the Beltrami identity. It is one of the most useful first integrals in elementary calculus of variations.

Some books write the same result with the opposite sign:

y′Fy′ − F = C1.

(27)

The two forms are equivalent because C₁ = −C. What matters is the constancy, not the sign convention.

6 Why a first integral is valuable

Euler–Lagrange is normally second order. A first integral has the schematic form

′ G (x,y,y ) = C.

(28)

If this relation can be solved for y′, then

′ y = Φ(x, y;C ).

(29)

This is a first-order differential equation. In favorable cases it can be separated:

dy ----------= dx, Φ(x, y;C)

(30)

or rearranged into another directly integrable form. Thus the first integral has reduced the differential order by one.

A first integral does not guarantee that the remaining first-order equation has an elementary closed-form solution. It may only reduce the problem to a quadrature. That reduction is still substantial.

7 Worked example 1: planar arc length revisited

Consider

∫ b∘ ------′2- J [y] = 1 + (y ) dx. a

(31)

The integrand is

′ ∘ --------- F (y) = 1 + (y′)2.

(32)

It contains neither y nor x explicitly, so both special structures are available.

7.1 Using the cyclic-variable form

Since F_y = 0,

y′ Fy′ = ∘----------= C. 1 + (y′)2

(33)

Square both sides:

′ 2 --(y-)---= C2. 1 + (y′)2

(34)

Therefore

(y′)2 = C2(1 + (y′)2),

(35)

(1 − C2 )(y′)2 = C2.

(36)

As long as |C| < 1,

y′ = constant.

(37)

Hence

-------------- |y = mx + c,| --------------

(38)

which is a straight line.

7.2 Using Beltrami instead

Because F_x = 0,

F − y′Fy′ = C.

(39)

Substitute the arc-length integrand:

F − y′F_y′	= −	(40)
	= .	(41)

Therefore

1 ∘----------= C, 1 + (y′)2

(42)

which again implies constant slope.

The two first integrals are not independent in this simple example; both encode the same straight-line geometry.

8 Worked example 2: a cyclic variable with explicit x dependence

Return to

∫ b1 ′2 J[y] = 2-a(x)(y) dx, a

(43)

where a(x) > 0 is known. Because F_y = 0,

′ Fy′ = a(x)y = C.

(44)

Thus

|----------| |y′ = -C--.| | a(x) | ------------

(45)

Integrating,

|---------∫-------------| | x d ξ | y(x ) = C a-(ξ) + D. | -------------------------

(46)

The constants C and D are then selected by the endpoint conditions.

Notice the distinction:

Fy = 0

(47)

gives a first integral even though

Fx ⁄= 0.

(48)

Therefore the Beltrami identity does not apply, but the cyclic-variable integral does.

9 Worked example 3: autonomous mechanics and energy

In analytical mechanics the independent variable is time t and the dependent variable is a generalized coordinate q(t). To keep the notation explicit, define the generalized velocity

v = dq-. dt

(49)

The integrand is the Lagrangian

F = L(q,v,t).

(50)

If the Lagrangian has no explicit time dependence,

Lt = 0,

(51)

Beltrami gives

L − vLv = C.

(52)

Define canonical momentum

p = L . v

(53)

Then

vp − L = − C.

(54)

The quantity

H = vp − L

(55)

is the Hamiltonian associated with the Lagrangian. Hence an autonomous Lagrangian gives

|--------------| -H-=--constant.-

(56)

For the standard one-degree-of-freedom mechanical Lagrangian

1 L = --mv2 − V (q), 2

(57)

we have

p = mv,

(58)

and therefore

H	= v(mv) −	(59)
	= mv² + V (q).	(60)

Thus

|------------------| |1mv2 + V (q ) = E,| -2------------------

(61)

which is conservation of mechanical energy for this class of systems [4, 5].

9.1 Order reduction by the energy integral

From

1mv2 + V (q ) = E, 2

(62)

we obtain

∘ -------------- 2 v = ± --(E − V (q )). m

(63)

Therefore

|------∘-----------------| | m------dq----- | |dt = ± 2 ∘ ---------.| ---------------E-−-V-(q)--

(64)

The second-order equation of motion has been reduced to a first-order equation, and then to a quadrature.

Figure. For an autonomous one-degree-of-freedom mechanical system, the Beltrami first integral becomes conservation of energy. Motion is possible only where E ≥ V (q); the intersections E = V (q) are turning points where v = 0.

10 Example 4: harmonic oscillator from the first integral

Let

V (q) = 1kq2. 2

(65)

Energy conservation gives

1mv2 + 1kq2 = E. 2 2

(66)

Suppose the turning-point amplitude is A. At q = A,

v = 0,

(67)

1 2 E = 2-kA .

(68)

Substitution gives

2 2 2 mv = k(A − q ).

(69)

Define

∘ --- ω = k-. m

(70)

Then

∘ -------- v = ± ω A2 − q2.

(71)

Since v = dq∕dt,

dq- ∘ --2----2 dt = ±ω A − q .

(72)

Separate variables:

∘---dq---- = ± ω dt. A2 − q2

(73)

Integrating,

q- arcsin A = ± ωt + ϕ,

(74)

which yields the familiar sinusoidal motion

|--------------------| |q(t) = A sin(ωt + ϕ).| ----------------------

(75)

The first integral did not merely confirm energy conservation; it solved the motion by one quadrature.

11 Worked example 5: a refractive-index invariant

A useful preview of variational optics is the functional

∫ ∘ ------′2- J[y] = n (y ) 1 + (y) dx,

(76)

where the refractive index depends on height y but not explicitly on x. The integrand is

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

(77)

Because F_x = 0, Beltrami gives

F − y′Fy′ = C.

(78)

Now

-----y′---- Fy′ = n(y)∘ ------′-2. 1 + (y )

(79)

Hence

F − y′F_y′	= n − n	(80)
	= .	(81)

Therefore

|----------------| | n(y) | |∘-------′2-= C. | ---1-+-(y)--------

(82)

Let 𝜃 be the angle between the ray tangent and the positive x axis. Then

y′ = tan 𝜃,

(83)

∘----1------= cos𝜃. 1 + (y ′)2

(84)

Thus

|--------------| n-(y)cos𝜃-=--C.-

(85)

If instead the ray angle α is measured from the vertical normal to the horizontal index layers, then α = π
2 − 𝜃 and

|--------------| n (y)sinα = C.| ----------------

(86)

This is the continuous-medium form of the same invariant underlying Snell’s law. CV16 will derive Fermat’s principle and optical ray equations in full.

Figure. When the optical integrand is independent of x, the Beltrami invariant is constant along the ray. The tangent angle changes as n(y) changes so that n cos 𝜃 remains fixed.

12 Do not confuse the two special cases

The two common first integrals come from different missing variables:


Structure	Euler–Lagrange consequence	First integral

F_y = 0	F_y′ = 0	F_y′ = C

F_x = 0	(F − y′F_y′) = 0	F − y′F_y′ = C

both F_y = 0 and F_x = 0	both reductions are available	the resulting constants may be algebraically related

It is therefore useful to inspect the integrand before differentiating anything. A ten-second structural check can save a page of second-order ODE algebra.

13 A compact derivation from Euler–Lagrange by multiplication

There is another way to see the Beltrami identity. Start from

Fy − -d-Fy′ = 0. dx

(87)

Multiply by y′:

′ ′-d- ′ Fyy − y dx Fy = 0.

(88)

If F_x = 0, the total derivative of F is

dF ′ ′′ --- = Fyy + Fy ′y . dx

(89)

Therefore

Fyy ′ = dF − Fy ′y ′′. dx

(90)

Substitute into the multiplied Euler–Lagrange equation:

dF-− Fy′y′′ − y ′ d-Fy′ = 0. dx dx

(91)

But by the product rule,

d-- ′ ′′ ′ d- dx (y Fy′) = y Fy′ + y dxFy ′.

(92)

Hence

dF-− d--(y′F ′) = 0, dx dx y

(93)

d ′ ---(F − y Fy′) = 0. dx

(94)

This derivation is algebraically equivalent to the du Bois–Reymond route but can be useful when first learning the identity.

14 Endpoint conditions still matter

A first integral reduces the interior differential equation; it does not remove the boundary data developed in CV05.

For example, suppose

Fy = 0

(95)

and therefore

Fy′ = C.

(96)

If the terminal value y(b) is free at a fixed b, CV05 gives the natural condition

Fy ′(b) = 0.

(97)

Since F_y′ is constant along the entire extremal, this forces

C = 0.

(98)

The endpoint condition can therefore determine the first-integral constant immediately.

This is a recurring pattern:

|--------------------------------------------------------------------| |interior symmetry gives the constant; boundary data select its value. | ---------------------------------------------------------------------

(99)

15 First integrals and symmetry: a preview of Noether

The two reductions in this article already contain the seed of Noether’s theorem.

If F_y = 0, shifting the dependent variable by a constant,

y −→ y + 𝜖,

(100)

does not change the integrand. The associated conserved quantity is

Fy′.

(101)

If F_x = 0, translating the independent variable,

x −→ x + 𝜖,

(102)

does not change the explicit form of the integrand. The associated first integral is

F − y′Fy′.

(103)

In mechanics these become, respectively, momentum-like and energy-like conservation laws. CV20 will replace these special observations by the general Noether theorem for continuous transformations.

16 Regularity and limitations

The derivations above use classical differentiability. A clean sufficient setting is

F ∈ C2, y ∈ C2. ∗

(104)

Then the chain rules and total derivatives used in the proofs are ordinary classical derivatives.

Several cautions are important.

A first integral is a necessary consequence of stationarity under the stated hypotheses. It does not by itself prove a minimum.
If F_y′y′ = 0, the Euler–Lagrange equation can be degenerate, and the usual interpretation as a second-order equation may fail.
Solving a first integral for y′ may require choosing branches. Boundary conditions and continuity determine which branch is admissible.
A first integral may reduce the problem only to an implicit quadrature. An elementary closed form is not guaranteed.
If a stationary curve has corners, additional matching conditions are needed; these are developed in CV10.

17 Common mistakes

Mistake: using Beltrami whenever F_y = 0. Beltrami requires F_x = 0; the cyclic-variable integral requires F_y = 0.
Mistake: concluding that a cyclic variable is constant. The conserved object is F_y′, not generally y.
Mistake: dropping the sign difference between F − y′F_y′ and y′F_y′ − F. Both are valid, but the integration constant changes sign.
Mistake: calling y′F_y′ − F “energy” in every variational problem. It is a Hamiltonian-like quantity; it becomes physical energy only under the appropriate mechanical interpretation.
Mistake: thinking a conserved quantity proves minimality. Conservation follows from stationarity and symmetry, not from a second-variation test.
Mistake: forgetting endpoint conditions after finding a first integral. The boundary data still determine the constants and may eliminate entire branches of solutions.

18 A practical first-integral checklist

Given

∫ b J[y] = F (x,y,y′)dx, a

(105)

use the following order of attack:

Inspect which variables appear explicitly in F.
If F_y = 0, write immediately
$Fy′ = C.$
If F_x = 0, write immediately
$F − y′Fy′ = C.$
If neither simplification applies, use the full Euler–Lagrange equation.
Solve the first-order relation for y′ when possible.
Integrate once more or reduce to a quadrature.
Apply fixed, natural, or transversality boundary conditions from CV05.
Keep the result labeled as a stationary candidate until a minimum or maximum classification has been established separately.

19 Summary

For

∫ b J[y] = F (x,y,y′)dx, a

(106)

Euler–Lagrange gives

F − -d-F ′ = 0. y dx y

(107)

If y is cyclic,

|------------------------| -Fy-=-0---=⇒----Fy-′ =-C.-

(108)

For any sufficiently smooth Euler–Lagrange extremal,

|--------------------| |d | |---(F − y′Fy′) = Fx.| -dx-------------------

(109)

If x is absent explicitly, this becomes the Beltrami identity

|------------------------------| Fx = 0 =⇒ F − y′Fy′ = C.| --------------------------------

(110)

These identities reduce the order of many variational differential equations and expose conservation structures long before the general Noether theorem is available. CV06E1 will practice the Beltrami identity on several nonlinear problems, including classical catenary-type reductions, while CV06E2 will focus on cyclic variables and conservation laws. CV07 next extends Euler–Lagrange to several dependent variables and coupled systems.

References

[1] I. M. Gelfand and S. V. Fomin, Calculus of Variations, Dover Publications, 2000.

[2] Bruce van Brunt, The Calculus of Variations, Springer, 2004.

[3] Robert Weinstock, Calculus of Variations with Applications to Physics and Engineering, Dover Publications, 1974.

[4] Cornelius Lanczos, The Variational Principles of Mechanics, 4th ed., Dover Publications, 1986.

[5] Herbert Goldstein, Charles Poole, and John Safko, Classical Mechanics, 3rd ed., Addison Wesley, 2002.

"Calculus of Variations: First Integrals and Special Forms of the Euler--Lagrange Equation" is owned by bloftin.
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