Calculus of Variations: Vector Trajectory Problems and Conserved Momentum

CV07 derived the Euler–Lagrange equations for several dependent variables, and CV07E1 used them for coupled matrix systems. This companion set changes the geometric viewpoint. The unknown is now a trajectory

( ) r(t) = x(t) y(t)

(1)

in the plane, or more generally a vector curve in configuration space. The central questions are:

how a scalar action produces a vector differential equation;
how fixed and free endpoint conditions act component by component;
when generalized momentum is a conserved vector;
how a spatial symmetry can conserve only selected momentum components;
how endpoint freedom produces a geometric transversality condition.

These are the trajectory-level forms of the vector Euler–Lagrange equations used throughout analytical mechanics [3, 4, 5]. The examples remain elementary enough that every stationary path can be found in closed form.

Figure. A vector trajectory and an admissible variation. With both endpoints fixed, the perturbation vector must vanish at the initial and final times, while it may point in arbitrary directions in the interior.

1 Formula sheet for this set

For

∫ t f ˙ S [r] = t0 L(t,r,r)dt,

(2)

the vector first variation is [1, 2]

∫ tf [∂L d ( ∂L ) ] [∂L ]tf δS = --- − -- --- ⋅ η dt + --- ⋅ η . t0 ∂r dt ∂r˙ ∂˙r t0

(3)

The generalized momentum vector is

∂L- p = ∂ ˙r.

(4)

The vector Euler–Lagrange equation is therefore

dp- ∂L- dt = ∂r .

(5)

If the Lagrangian has no dependence on the position vector,

∂L ---= 0, ∂r

(6)

then

dp-= 0, dt

(7)

so the entire generalized momentum vector is conserved. If only one coordinate is absent from the Lagrangian, only the corresponding momentum component is guaranteed to be conserved. This is the finite-dimensional preview of the symmetry ideas developed systematically in CV20.

Part I: Exercises

Exercise 1: vector Dirichlet energy and the straight path

Let

1 ∫ 1 J [r] = -- ∥r ′(s)∥2ds, 2 0

(8)

where

r(0) = A, r(1) = B.

(9)

Derive the vector Euler–Lagrange equation.
Solve the resulting boundary-value problem.
Show directly that the stationary path is also the unique global minimum of this quadratic functional.

Exercise 2: a free particle and conservation of the momentum vector

For a particle of mass $m$ , consider

∫ tf1 S [r] = --m ∥˙r∥2dt t0 2

(10)

with fixed endpoint positions $r0$ and $rf$ .

Compute the generalized momentum vector.
Use Euler–Lagrange to prove that this vector is conserved.
Solve for the stationary trajectory joining the two endpoint events.
Interpret the result geometrically.

Exercise 3: uniform gravity with two fixed endpoint events

Take

1 2 2 L = --m (˙x + ˙y ) − mgy 2

(11)

for $0 ≤ t ≤ T$ , with

(x(0),y(0)) = (x0,y0), (x (T ),y(T )) = (xf ,yf).

(12)

Derive the two Euler–Lagrange equations.
Identify which momentum component is conserved and explain why.
Solve the boundary-value problem explicitly.
Show that the spatial trajectory is parabolic when $xf ⁄= x0$ .

Exercise 4: one cyclic spatial coordinate

Let

1- 2 2 L = 2m (˙x + ˙y ) − V (y ).

(13)

Derive the Euler–Lagrange equations.
Prove that $px$ is conserved.
Explain why $p y$ is not generally conserved.
State the additional condition under which the full vector $p$ is conserved.

Exercise 5: a translation-invariant anisotropic kinetic energy

Let $q (t) ∈ ℝn$ and

L = 1q˙T M q˙, 2

(14)

where $M$ is a constant symmetric positive-definite matrix. The endpoint positions are fixed at $q 0$ and $q f$ over an elapsed time $T$ .

Find the generalized momentum vector.
Prove that it is constant.
Find the stationary path.
Express the constant momentum in terms of $q0$ , $qf$ , $T$ , and $M$ .

Exercise 6: one fixed endpoint component and one free component

Suppose the initial position is fixed. At $tf$ , the final horizontal coordinate is prescribed,

x (tf) = X,

(15)

but $y (tf)$ is free. The final time is fixed.

State the allowed endpoint variation $η (tf)$ .
Use the boundary term in $δS$ to derive the natural endpoint condition.
For the standard mechanical Lagrangian
$L = 1m (˙x2 + ˙y2) − V(x, y), 2$ (16)

interpret the endpoint condition physically.

Exercise 7: endpoint constrained to a target curve

The initial position and final time are fixed, but the final point may slide along a smooth target curve $C$ . Let $τ$ be a tangent vector to $C$ at the terminal point.

Show that the allowed endpoint variation is tangent to $C$ .
Derive
$p ⋅ τ = 0. f$ (17)
If the target curve is represented by $G (x, y) = 0$ , show that
$pf = λ∇G$ (18)

for some scalar $λ$ .
For a free particle, explain the geometric meaning of the condition.

Exercise 8: a terminal endpoint cost

Let

∫ tf J [r] = L (t,r,r˙) dt + Φ (r(tf)), t0

(19)

where the initial position is fixed and the final position is free.

Derive the terminal condition relating $pf$ to $∇ Φ$ .
For
$κ Φ (rf) = -∥rf − rd∥2, 2$ (20)

write the terminal condition explicitly.
Explain how the endpoint penalty changes the natural condition $pf = 0$ .

Exercise 9: momentum-vector conservation as a symmetry test

For a general planar Lagrangian

L = L (t,x, y, ˙x, ˙y),

(21)

classify the momentum conservation implied by each case:

$L$ is independent of $x$ but depends on $y$ .
$L$ is independent of $y$ but depends on $x$ .
$L$ is independent of both $x$ and $y$ .
$L$ depends on $x$ and $y$ only through a central potential $∘ -2----2- V( x + y )$ . Is the linear momentum vector generally conserved?

Part II: Complete worked solutions

Solution 1: vector Dirichlet energy and the straight path

The integrand is

1 F (r,r′) = --r′ ⋅ r′. 2

(22)

Therefore

∂F--= 0, ∂F--= r′. ∂r ∂r ′

(23)

The vector Euler–Lagrange equation gives

d ′ ds-r = 0,

(24)

′′ r = 0.

(25)

Integrating twice,

r(s) = C0 + sC1.

(26)

The endpoint conditions imply

C0 = A, C1 = B − A.

(27)

Hence

|-----------------------| r∗(s) = (1 − s)A + sB. | ------------------------

(28)

This is the straight segment joining the two endpoint vectors.

To prove global minimality directly, write any admissible competitor as

r = r + u, ∗

(29)

where

u (0) = u(1) = 0.

(30)

Then

J[r]	= ∫ ₀¹∥r_∗′ + u′∥²ds	(31)
	= J[r_∗] + ∫ ₀¹r_∗′⋅ u′ds + ∫ ₀¹∥u′∥²ds.	(32)

Because $′ r∗ = B − A$ is constant,

∫ 1 ′ ′ r∗ ⋅ u ds = (B − A ) ⋅ [u(1) − u(0)] = 0. 0

(33)

Thus

1 ∫ 1 J[r] − J [r∗] =-- ∥u ′∥2ds ≥ 0. 2 0

(34)

Equality requires $u′ = 0$ , and the endpoint conditions then force $u = 0$ . Therefore the straight path is the unique global minimizer.

Solution 2: a free particle and conservation of the momentum vector

For

L = 1-mr˙⋅ ˙r, 2

(35)

the generalized momentum is

|--------| -p-=-m-˙r.-

(36)

The Lagrangian contains no $r$ , so

∂L-= 0. ∂r

(37)

Euler–Lagrange therefore gives

p˙ = 0.

(38)

Hence

|p-=-constant.-| ---------------|

(39)

For a constant mass, $˙r$ is constant as well. Therefore

-t −-t0- r(t) = r0 + t − t (rf − r0). f 0

(40)

The stationary path in configuration space is a straight line traversed at constant velocity.

Figure. For a free particle the Lagrangian is invariant under spatial translation, so both planar coordinates are cyclic and the entire momentum vector is constant. The trajectory between fixed endpoint events is therefore straight with uniform velocity.

This is a simple but important example of a vector conservation law. The two component equations

˙p = 0, p˙ = 0 x y

(41)

are equivalent to the single vector statement $˙p = 0$ .

Solution 3: uniform gravity with two fixed endpoint events

The Lagrangian is

1 L = -m (x˙2 + y˙2) − mgy. 2

(42)

For $x$ ,

∂L ∂L --- = 0, --- = m x˙. ∂x ∂ ˙x

(43)

Thus

m ¨x = 0.

(44)

For $y$ ,

∂L- ∂L- ∂y = − mg, ∂ ˙y = m y˙,

(45)

m ¨y = − mg,

(46)

|------------------| ¨x-=--0,----¨y-=-−-g.-

(47)

Since $x$ is cyclic,

|--------------------| px-=--m-˙x =-constant.-

(48)

But $L$ depends explicitly on $y$ , so $py = m y˙$ is not conserved. Indeed,

p˙y = − mg.

(49)

The horizontal solution satisfying the endpoints is

x(t) = x + xf-−-x0-t. 0 T

(50)

The vertical solution has the form

1- 2 y(t) = y0 + vy0t − 2 gt.

(51)

Imposing $y(T ) = y f$ gives

yf − y0 1 vy0 = ------- + --gT. T 2

(52)

Therefore

|-----------(---------------)----------| | yf −-y0 1- 1- 2 | y (t) = y0 + T + 2gT t − 2gt .| ----------------------------------------

(53)

If $x ⁄= x f 0$ , then

x − x0 t = T--------. xf − x0

(54)

Substitution into $y(t)$ produces a quadratic polynomial in $x$ , so the spatial trajectory is a parabola.

Figure. A stationary mechanical trajectory in a uniform gravitational field. Spatial translation symmetry remains in the horizontal direction, giving constant $px$ , while gravity changes the vertical momentum.

This exercise illustrates why conservation should be checked component by component before it is promoted to a vector statement.

Solution 4: one cyclic spatial coordinate

For

1 2 2 L = -m (˙x + ˙y ) − V (y ), 2

(55)

the $x$ equation is

d-(m ˙x) = 0, dt

(56)

|--------------------| px = m ˙x = constant.| ----------------------

(57)

The $y$ equation is

-d ′ dt(m y˙) − [− V (y)] = 0,

(58)

hence

|--------------| |m ¨y = − V′(y).| ----------------

(59)

Thus

p˙y = − V ′(y),

(60)

which is generally nonzero.

The full vector momentum

(x˙) p = m y˙

(61)

is conserved only when the Lagrangian is independent of both $x$ and $y$ . For this example that requires

V′(y) = 0,

(62)

so the potential must be constant over the region of interest.

Solution 5: a translation-invariant anisotropic kinetic energy

The Lagrangian is

L = 1q˙T M q˙. 2

(63)

Because $M$ is symmetric,

|----------------| | ∂L | |p = ---= M q˙. | ------∂˙q---------

(64)

There is no explicit dependence on $q$ , so

p˙ = 0.

(65)

Since $M$ is constant and nonsingular,

M ¨q = 0

(66)

implies

q¨= 0.

(67)

For an elapsed time $T$ , the stationary path is

|------------t--------------------------| q (t) = q0 + --(qf − q0 ), 0 ≤ t ≤ T. | ------------T----------------------------

(68)

Its constant velocity is

q˙= qf-−-q0, T

(69)

|-------------------| | 1- | p = T M (qf − q0 ). | ---------------------

(70)

The path is still affine in configuration space. The matrix $M$ changes the relationship between velocity and momentum, not the fact that both are constant for this translation-invariant quadratic Lagrangian.

Solution 6: one fixed endpoint component and one free component

At the initial time,

η(t0) = 0.

(71)

At the final time, $x(tf) = X$ is fixed, so

ηx(tf) = 0.

(72)

The final $y$ value is free, so $ηy(tf)$ is arbitrary. Thus

( ) 0 η(tf) = ηy(tf) .

(73)

After the interior Euler–Lagrange equations are satisfied, the remaining endpoint term is

pf ⋅ ηf = px (tf) 0 + py (tf)ηy(tf).

(74)

Because $ηy(tf)$ is arbitrary,

|----------| py(tf)-=-0.-

(75)

For

L = 1m (˙x2 + ˙y2) − V(x, y), 2

(76)

we have $py = m ˙y$ , hence

|----------| |˙y(t ) = 0.| ----f------

(77)

The final velocity has no component along the direction in which the endpoint is free to move. This is the componentwise natural boundary condition derived abstractly in CV05.

Solution 7: endpoint constrained to a target curve

Let $r f$ lie on a smooth target curve $C$ . The endpoint may move along the curve but not away from it. Therefore an admissible first-order endpoint displacement has the form

ηf = ατ ,

(78)

where $τ$ is tangent to $C$ and $α$ is arbitrary.

The endpoint contribution to the first variation is

pf ⋅ η = α pf ⋅ τ . f

(79)

Stationarity for every allowed $α$ requires

|----------| pf-⋅-τ-=-0.-

(80)

Thus the terminal momentum is normal to the target curve.

If the curve is defined by

G (x, y) = 0,

(81)

then $∇G$ is normal to the curve. Consequently

|--------------| pf = λ∇G (rf)| ----------------

(82)

for some scalar $λ$ .

Figure. At a fixed final time, a terminal point constrained to a curve may vary only along the curve tangent. The variational endpoint term therefore forces the terminal generalized momentum to be normal to every allowed endpoint motion.

For a free particle, $p = m ˙r$ , so the terminal velocity is normal to $C$ . In the Euclidean kinetic-energy problem this reproduces the familiar orthogonality of a shortest path to a freely chosen point on a target curve.

Solution 8: a terminal endpoint cost

Consider

∫ tf J[r] = L dt + Φ (rf). t0

(83)

The variation of the terminal cost is

δΦ = ∇ Φ (rf) ⋅ ηf .

(84)

After imposing the interior Euler–Lagrange equations and the fixed initial endpoint, the terminal part of the first variation is

[pf + ∇ Φ (rf)] ⋅ ηf.

(85)

Because the final position is completely free, all components of $ηf$ are arbitrary. Hence

|------------------| |pf + ∇ Φ (rf ) = 0.| -------------------

(86)

For the quadratic terminal penalty

Φ (rf) = κ∥rf − rd∥2, 2

(87)

we have

∇ Φ = κ(rf − rd).

(88)

Therefore

|------------------| |p = − κ(r − r ).| --f--------f----d---

(89)

Without a terminal cost, a completely free endpoint would give $pf = 0$ . The endpoint penalty replaces that zero-momentum condition by a balance between terminal momentum and the gradient of the endpoint objective. This structure later reappears in optimal control.

Solution 9: momentum-vector conservation as a symmetry test

The component Euler–Lagrange equations are

∂L- ∂L- p˙x = ∂x , p˙y = ∂y.

(90)

Therefore:

If $L$ is independent of $x$ , then
$px = constant,$ (91)

but $py$ need not be constant.
If $L$ is independent of $y$ , then
$p = constant, y$ (92)

but $px$ need not be constant.
If $L$ is independent of both $x$ and $y$ , then
$|--------------| -p-=-constant.-|$ (93)

The full translation group acts as a symmetry of the Lagrangian.
For a central potential
$∘ -------- V = V (r), r = x2 + y2,$ (94)

the Lagrangian depends on position, so the linear momentum vector is not generally conserved. The force points radially and changes the momentum vector. Rotational symmetry instead leads to angular-momentum conservation, a result developed systematically through Noether’s theorem in CV20 [4, 5].

2 A compact endpoint-condition map for vector trajectories

For a stationary vector trajectory, the boundary contribution at a fixed final time is

pf ⋅ ηf .

(95)

The allowed endpoint motion determines the terminal condition:


Endpoint freedom	Allowed variation	Stationarity condition

Fully fixed	$ηf = 0$	no additional natural condition
All components free	arbitrary $ηf$	$pf = 0$
Only $yf$ free	$(0, ηy)T$	$py(tf) = 0$
Constrained to curve $C$	tangent to $C$	$pf ⊥ C$
Free with terminal cost $Φ$	arbitrary $ηf$	$pf + ∇ Φ = 0$

The important lesson is that a vector endpoint condition is not usually imposed all at once. Each free direction contributes one scalar transversality condition.

3 Common mistakes

Mistake: treating $p$ as conserved whenever one coordinate is cyclic. A single cyclic coordinate conserves one momentum component, not necessarily the entire vector.
Mistake: forgetting that fixed endpoint values make the corresponding components of $η$ vanish.
Mistake: setting every terminal momentum component to zero when the endpoint is constrained to a curve. Only momentum components along allowed endpoint directions must vanish.
Mistake: assuming that a conserved generalized momentum is always $m ˙r$ . For a general Lagrangian it is $∂L ∕∂˙r$ .
Mistake: confusing the stationary mechanical trajectory with a shortest spatial path. Hamilton’s action and Euclidean path length are different functionals.
Mistake: using Euler–Lagrange stationarity alone as a proof of global minimality. Exercise 1 admits a direct positivity proof, but general mechanical actions need not be minima.

4 Summary

For a vector trajectory $r(t)$ , the Euler–Lagrange equation may be written

|--(----)------------| |d ∂L ∂L | |-- --- − --- = 0.| -dt--∂-˙r-----∂r-------

(96)

The generalized momentum vector is

p = ∂L-. ∂ ˙r

(97)

If every spatial coordinate is cyclic, then

|------| p˙-=-0.-

(98)

If only selected coordinates are cyclic, only the corresponding momentum components are conserved.

At a fixed terminal time, endpoint freedom is encoded in

pf ⋅ ηf = 0.

(99)

A completely free endpoint gives $pf = 0$ ; a point constrained to a curve forces $pf$ normal to that curve; and a terminal cost changes the condition to

p + ∇ Φ = 0. f

(100)

CV07E2 therefore ties together vector Euler–Lagrange equations, trajectory geometry, conservation laws, and the endpoint conditions introduced in CV05. The next main lesson, CV08, increases the differential order of the functional and shows how repeated integration by parts creates higher-order Euler–Lagrange equations and additional boundary terms.

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

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

[5] V. I. Arnold, Mathematical Methods of Classical Mechanics, 2nd ed., Springer, 1989.

"Calculus of Variations: Vector Trajectory Problems and Conserved Momentum" is owned by bloftin.
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