Calculus of Variations: Hamilton's Principle and Classical Mechanics - Exercises

(Example)

Calculus of Variations: Hamilton’s Principle and Classical Mechanics - Exercises and Worked Solutions

CV14 introduced Hamilton’s principle as the mechanical specialization of the calculus of variations:

∫ t 2 ⋅ S [q] = t L (q,q,t)dt, δS = 0. 1

(1)

The central message was that the physical trajectory is a stationary trajectory of the action, not automatically a minimizing trajectory. This exercise set develops that statement computationally. The problems begin with the free particle and uniform gravity, then move through the harmonic oscillator, pendulum, generalized coordinates, cyclic coordinates, conserved energy, total-derivative changes of the Lagrangian, and natural endpoint conditions. These are standard themes in analytical mechanics [1, 2, 3, 4] and they connect directly to the variational machinery developed earlier in this series [5].

Figure. A practical workflow for Hamilton-principle exercises. The coordinates and energy model define L; the Euler–Lagrange equations determine stationary trajectories; symmetries supply first integrals; and CV11–CV13 determine whether a stationary path is actually minimizing.

1 Exercises

Exercise 1: free particle with fixed endpoints

A particle of mass m moves in one dimension with

L(q,q⋅) = 1-m ⋅q2. 2

(2)

The endpoint conditions are

q(0) = 0, q(T ) = a, T > 0.

(3)

Derive the Euler–Lagrange equation.
Solve the boundary-value problem for the stationary path q_∗(t).
Evaluate the stationary action S[q_∗].
Let q = q_∗ + η with η(0) = η(T) = 0. Show directly that
$∫ T S[q] − S [q∗] = m ⋅η2dt. 2 0$ (4)
Conclude that q_∗ is the unique global minimizer of the action for these fixed endpoints.

Figure. The straight worldline in the (t,q) plane is stationary. For the free-particle quadratic action, every nonzero fixed-endpoint perturbation adds the positive amount 1
2 m∫ ⋅
η ²dt.

Exercise 2: a projectile between prescribed spacetime endpoints

A particle moves in the vertical plane under uniform gravity. Use

L = 1m (x⋅2 + y⋅2) − mgy. 2

(5)

The particle must travel from

(x,y ) = (0, 0) at t = 0

(6)

(x,y) = (X, Y ) at t = T.

(7)

Derive the two Euler–Lagrange equations.
Solve the fixed-endpoint boundary-value problem.
Determine the required initial velocity components.
Verify that the resulting curve is a parabola in the x-y plane when X≠0.
Explain why the varied paths in Hamilton’s principle need not themselves satisfy the projectile equations.

Figure. Uniform gravity with fixed spacetime endpoints. The endpoint conditions determine the initial velocity required for the stationary projectile trajectory.

Exercise 3: harmonic oscillator boundary-value problem and critical time

For

∘ --- ⋅ 1 ⋅2 1 k L(q,q) = --m q − -kq2, ω = --, 2 2 m

(8)

impose

q(0) = 0, q (T ) = Q.

(9)

Derive the Euler–Lagrange equation.
For sin(ωT)≠0, find the unique stationary trajectory.
What happens if Q≠0 but sin(ωT) = 0?
Now set Q = 0. Show that at
$nπ- T = ω , n = 1, 2,...,$ (10)

there are nonzero stationary trajectories satisfying both endpoint conditions.
For the stationary path q_∗ = 0, compute the second variation and use the trial direction
$( ) η(t) = sin πt- T$ (11)

to identify the first critical time.
Classify the three regimes T < π∕ω, T = π∕ω, and T > π∕ω in the language of CV11–CV12.

Figure. The first harmonic-oscillator conjugate time is π∕ω. The same stationary equation persists on longer intervals, but the minimizing character changes.

Exercise 4: the simple pendulum

A pendulum of mass m and length ℓ is described by the angle 𝜃 from the downward vertical.

Compute the kinetic energy and gravitational potential energy using
$V (𝜃) = mg ℓ(1 − cos𝜃).$ (12)
Form the Lagrangian.
Derive the exact pendulum equation.
Because the Lagrangian is autonomous, compute the conserved energy function
$⋅∂L HL = 𝜃--⋅ − L. ∂𝜃$ (13)
Linearize the equation for |𝜃|≪ 1 and identify the small-angle frequency.
Explain why the linearized equation is an approximation to the dynamics, whereas the variational derivation of the exact pendulum equation is not.

Exercise 5: a central force and a cyclic angle

For planar motion in polar coordinates, take

2 ⋅2 L = 1m (r⋅ + r2𝜃 ) − V(r). 2

(14)

Show that 𝜃 is cyclic.
Derive the conserved generalized momentum p_𝜃.
Derive the radial Euler–Lagrange equation.
Eliminate in favor of the constant angular momentum ℓ.
Show that the radial motion can be written using the effective potential
$ℓ2 Veff (r) = V (r) + ----2. 2mr$ (15)
Explain physically why the ℓ²∕(2mr²) term acts as a centrifugal barrier.

Figure. Rotational symmetry makes 𝜃 cyclic. Conservation of p_𝜃 = mr² ⋅
𝜃 reduces the two-coordinate problem to one radial equation with an effective potential.

Exercise 6: deriving the energy identity for an arbitrary Lagrangian

Let

L = L(q,q⋅,t)

(16)

and define

HL = q⋅L ⋅− L. q

(17)

Differentiate H_L with respect to time.
Use the Euler–Lagrange equation to show
$|------------| |dH | |---L-= − Lt.| --dt----------$ (18)
State the conservation law when L_t = 0.
For
$1- ⋅2 L = 2 m q − V (q),$ (19)

show that H_L = T + V .
Explain why this derivation is a first integral of the Euler–Lagrange equation rather than an independent postulate of energy conservation.

Exercise 7: a generalized coordinate with coordinate-dependent inertia

Consider

L (q, ⋅q) = 1M (q)q⋅2 − V (q), M (q ) > 0. 2

(20)

Compute the generalized momentum p.
Derive the Euler–Lagrange equation and show that it can be written
$M (q)⋅q⋅+ 1-M ′(q)q⋅2 + V ′(q) = 0. 2$ (21)
Compute H_L and show that it equals
$1- ⋅2 2 M (q)q + V (q ).$ (22)
Explain why p = M(q) is generally not conserved even though energy is conserved.
Identify the special condition under which p would be conserved.

Exercise 8: adding a total time derivative to the Lagrangian

Let

⋅ ⋅ d ^L(q,q,t) = L (q,q, t) + --F (q,t). dt

(23)

Expand dF∕dt using the chain rule.
Show directly that L and L produce the same Euler–Lagrange equation.
Show that their actions differ by the endpoint quantity
$^S − S = F(q(t2),t2) − F (q(t1),t1).$ (24)
Explain why the two actions therefore have the same stationary paths when the endpoints are fixed.
Show that the canonical momentum changes according to
$^p = p + Fq.$ (25)
Explain why this example is an early warning that canonical momentum is representation-dependent even when the physical trajectory is unchanged.

Exercise 9: free terminal position and the momentum boundary term

Consider

∫ t2 ⋅ S[q] = L (q,q,t)dt, t1

(26)

with fixed initial position q(t₁), fixed endpoint times, but free terminal position q(t₂).

Repeat the first-variation calculation without imposing η(t₂) = 0.
Show that stationarity requires the Euler–Lagrange equation in the interior and the natural terminal condition
$|------------------| |p(t2) = L ⋅(t2) = 0.| ----------q---------$ (27)
Now add a terminal cost
$Φ(q(t2))$ (28)

to the action. Derive the modified terminal condition

$|---------------------| p(t2) +-Φq(q(t2))-=-0.--$ (29)
Explain how this connects the endpoint theory of CV05 with generalized momentum in mechanics.

2 Worked solutions

Solution 1: free particle with fixed endpoints

The action is

m ∫ T ⋅2 S [q ] =-- q dt. 2 0

(30)

The Euler–Lagrange equation is

d-(m ⋅q) = 0, dt

(31)

⋅⋅q = 0.

(32)

Hence

q (t) = At + B. ∗

(33)

The boundary conditions give B = 0 and AT = a, so

|------------| | a- | -q∗(t) =-T-t.-|

(34)

The stationary action is

m ∫ T ( a)2 ma2 S [q∗] = -- -- dt = ----. 2 0 T 2T

(35)

Now write

q = q∗ + η, η(0) = η(T ) = 0.

(36)

Then

m ∫ T S [q] = -- (q⋅∗ + η⋅)2dt. 2 0

(37)

Expanding the square gives

∫ ∫ T ⋅ ⋅ m- T ⋅2 S[q] = S [q∗] + m q∗η dt + 2 η dt. 0 0

(38)

Because _∗ = a∕T is constant,

∫ T a ⋅q∗⋅η dt = --[η]T0 = 0. 0 T

(39)

Therefore

|------------------------------| | m ∫ T ⋅2 | |S[q] − S [q∗] =-- η dt ≥ 0.| ----------------2--0------------

(40)

Equality requires = 0, so η is constant. The endpoint conditions then force η ≡ 0. Thus the straight worldline is the unique global minimizer.

Solution 2: a projectile between prescribed spacetime endpoints

For x,

Lx = 0, L ⋅= mx⋅, x

(41)

|⋅⋅-----| -x-=-0.-

(42)

For y,

⋅ Ly = − mg, Ly⋅= m y,

(43)

|--------| |⋅⋅y = − g.| ----------

(44)

Integrating,

x(t) = vx0t,

(45)

because x(0) = 0. Since x(T) = X,

|---------| |-----------| |v = X--, |x(t) = X-t.| --x0---T--| --------T----

(46)

Similarly,

1 2 y (t) = vy0t − 2gt .

(47)

The terminal condition gives

1 Y = vy0T − -gT 2, 2

(48)

hence

|----------------| |v = Y- + 1gT | --y0---T----2----|

(49)

and

|----------------------------| | ( Y 1 ) 1 | y (t) = -- + -gT t − -gt2.| ---------T----2---------2-----

(50)

When X≠0,

t = T-x, X

(51)

( Y gT 2) gT2 y(x) = ---+ ---- x − ----2x2, X 2X 2X

(52)

which is a parabola.

The varied paths used in the action comparison are not required to obey ⋅⋅
x = 0 and ⋅y⋅ = −g. Those equations are precisely what stationarity is used to derive. Requiring every varied path to satisfy them would remove the freedom needed by the variational argument.

Solution 3: harmonic oscillator boundary-value problem and critical time

The Euler–Lagrange equation is

d ⋅ -- (m q) + kq = 0, dt

(53)

|------------| |⋅⋅q + ω2q = 0 , ω2 = k-. ------------- m

(54)

The general solution is

q(t) = A sin ωt + B cosωt.

(55)

Since q(0) = 0, B = 0. Therefore

Q = A sin ωT.

(56)

If sin ωT≠0,

|----------------| | sinωt--| q∗(t) = Q sin ωT .| ------------------

(57)

If Q≠0 and sin ωT = 0, the equation Q = A sin ωT cannot be satisfied. There is no classical stationary trajectory with those fixed endpoint data and that prescribed travel time.

If Q = 0 and

T = n-π, ω

(58)

then

qA(t) = A sin ωt

(59)

satisfies both endpoints for every real A. The boundary-value problem is therefore degenerate.

Along q_∗ = 0, the action is quadratic, so the second variation is

|--------------------------------| | ∫ T ( 2 ) | δ2S [0;η] = m ⋅η − ω2η2 dt.| ---------------0------------------

(60)

Take

( πt) η(t) = sin --- . T

(61)

Using

∫ T ∫ T 2 η2dt = T-, η⋅2dt = π--, 0 2 0 2T

(62)

we obtain

( ) 2 m- π2- 2 δ S = 2 T − ω T .

(63)

The first sign change occurs at

|-----π--| |Tc = --.| ------ω--|

(64)

Thus:

|------------------------------------------------------------| |T < π∕ ω : positive second variation; minimizing regime, | |T = π∕ ω : null Jacobi direction; first conjugate endpoint, | |T > π∕ ω : negative direction exists; not a minimum. | --------------------------------------------------------------

(65)

This is exactly the CV11–CV12 distinction between stationarity and conjugate-point loss of minimizing character.

Solution 4: the simple pendulum

The bob speed is

⋅ v = ℓ𝜃,

(66)

1 ⋅2 T = -m ℓ2𝜃 . 2

(67)

With the downward vertical as zero potential,

V = mg ℓ(1 − cos 𝜃).

(68)

Hence

|------------------------------| | 1 2⋅2 | |L = -m ℓ 𝜃 − mg ℓ(1 − cos 𝜃).| ------2------------------------

(69)

Now

⋅ L 𝜃 = − mg ℓsin𝜃, L ⋅= m ℓ2𝜃. 𝜃

(70)

Therefore

⋅⋅ m ℓ2𝜃 + mg ℓsin𝜃 = 0,

(71)

⋅⋅--g-----------| 𝜃 + --sin𝜃 = 0. | ----ℓ------------

(72)

The generalized momentum is

p = m ℓ2⋅𝜃. 𝜃

(73)

Because L has no explicit time dependence,

H = ⋅𝜃p − L. L 𝜃

(74)

Substituting the pendulum Lagrangian,

[ ] 2 ⋅2 1 2⋅2 HL = m ℓ 𝜃 − --m ℓ 𝜃 − mg ℓ(1 − cos 𝜃) . 2

(75)

|------------------------------------| | 1 2⋅2 | |HL = -m ℓ 𝜃 + mg ℓ(1 − cos𝜃) = E. | -------2------------------------------

(76)

For small angles,

sin𝜃 ≈ 𝜃,

(77)

and the equation becomes

⋅⋅ g 𝜃 + ℓ𝜃 = 0.

(78)

Therefore

|-----∘----| | g-| |ω0 = ℓ.| ------------

(79)

Hamilton’s principle produced the exact nonlinear equation. The approximation enters only afterward when sin 𝜃 is replaced by 𝜃.

Solution 5: a central force and a cyclic angle

Since L contains 𝜃 only through ⋅
𝜃 ,

L 𝜃 = 0.

(80)

Therefore

d-L ⋅= 0. dt 𝜃

(81)

But

L ⋅= mr2 𝜃⋅, 𝜃

(82)

|--------------------------| | ⋅ | -p𝜃 =-mr2-𝜃 =-ℓ =-constant.-

(83)

For r,

⋅ L ⋅r = m r,

(84)

and

⋅2 Lr = mr 𝜃 − V ′(r).

(85)

Hence

|------------------------| | ⋅⋅ ⋅2 ′ | -m-r −-mr-𝜃-+-V--(r) =-0.|

(86)

Using

⋅ 𝜃 = --ℓ- , mr2

(87)

we obtain

⋅⋅ -ℓ2- ′ m r = mr3 − V (r).

(88)

Define

|-------------------2--| Veff (r) = V (r) + --ℓ--.| -----------------2mr2---

(89)

Then

dVe-ff ′ -ℓ2- dr = V (r) − mr3 ,

(90)

and the radial equation becomes

|--------------| | ⋅⋅ dVeff | |m r = − -dr--.| ---------------

(91)

For ℓ≠0, the effective term

2 --ℓ--- 2mr2

(92)

diverges as r → 0. It therefore produces an angular-momentum barrier against reaching the origin unless the original potential is sufficiently singular to overcome it.

Solution 6: deriving the energy identity for an arbitrary Lagrangian

Start from

⋅ HL = qL ⋅q − L.

(93)

Differentiate:

dH ⋅⋅ ⋅d dL ---L-= qL ⋅q + q--Lq⋅− ---. dt dt dt

(94)

The total derivative of L is

dL- ⋅ ⋅⋅⋅ dt = Lqq + L qq + Lt.

(95)

Substitution gives

dHL ⋅( d ) -----= q --L⋅q − Lq − Lt. dt dt

(96)

Along an Euler–Lagrange trajectory, the parenthesis vanishes. Therefore

|------------| |dHL-- | | dt = − Lt.| --------------

(97)

If L_t = 0,

|----------------| -HL--=-constant.-|

(98)

For

1 ⋅2 L = --m q − V (q), 2

(99)

we have

⋅ L ⋅q = m q,

(100)

( ) HL = m q⋅2 − 1m q⋅2 − V = 1m ⋅q2 + V (q ). 2 2

(101)

Thus ordinary mechanical energy appears as a first integral of the Euler–Lagrange equation when the Lagrangian is autonomous.

Solution 7: a generalized coordinate with coordinate-dependent inertia

The generalized momentum is

|----------------| | ⋅ | p-=--L⋅q-=-M-(q)q.-

(102)

Next,

d d ⋅ ⋅⋅ ′⋅2 --L⋅q = -- [M (q )q] = M q + M q . dt dt

(103)

Also,

1 2 Lq = -M ′(q)⋅q − V ′(q). 2

(104)

The Euler–Lagrange equation therefore gives

( ) ⋅⋅ ′⋅2 1- ′⋅2 ′ M q + M q − 2M q − V = 0,

(105)

hence

|-----⋅⋅--1-------⋅2-------------| |M (q)q + --M ′(q)q + V ′(q) = 0. | ----------2----------------------|

(106)

The energy function is

[ ] ⋅ ⋅ 1- ⋅2 1- ⋅2 HL = q[M (q)q] − 2M (q)q − V (q) = 2M (q)q + V (q).

(107)

Because M and V can depend on q, the coordinate q is not generally cyclic. Therefore p = M is not generally conserved. It would be conserved only if

Lq = 0,

(108)

that is, if q were cyclic.

Solution 8: adding a total time derivative to the Lagrangian

By the chain rule,

dF ⋅ --- = Fq q + Ft. dt

(109)

Therefore

⋅ ^L = L + Fqq + Ft.

(110)

Compute

^Lq⋅= L ⋅q + Fq.

(111)

Then

d d ⋅ --^L ⋅q = --Lq⋅+ Fqqq + Fqt. dt dt

(112)

Also,

^Lq = Lq + Fqq⋅q + Ftq.

(113)

Since mixed partial derivatives agree,

F = F , qt tq

(114)

we get

d-^ ^ d- dtL ⋅q − Lq = dtL ⋅q − Lq.

(115)

Thus L and L produce exactly the same Euler–Lagrange equation.

At the action level,

∫ t2 S^− S = dF- dt = F (q(t),t ) − F(q(t ),t). t1 dt 2 2 1 1

(116)

For fixed endpoint data this is a fixed number, so the two actions have the same stationary paths.

However,

|----------------| p^= ^L⋅q = p + Fq.| ------------------

(117)

Thus a total derivative leaves the motion unchanged while shifting the canonical momentum. This is one reason canonical quantities must be interpreted with care.

Solution 9: free terminal position and the momentum boundary term

Let

q = q + 𝜀η, 𝜀 ∗

(118)

with

η(t1) = 0,

(119)

but no restriction on η(t₂).

The first variation is

∫ t2 ( ⋅) δS = Lqη + L ⋅qη dt. t1

(120)

Integrate by parts:

[ ] ∫ t [ ] t2 2 d- δS = Lq⋅η t1 + t Lq − dtL⋅q ηdt. 1

(121)

The initial boundary contribution vanishes because η(t₁) = 0. Stationarity for arbitrary interior variations gives the Euler–Lagrange equation. The remaining terminal term is

L ⋅q(t2)η(t2).

(122)

Because the terminal position is free, η(t₂) is arbitrary. Hence

|------------------| |p(t2) = L ⋅q(t2) = 0.| --------------------

(123)

Now consider

∫ t2 𝒮[q] = Ldt + Φ (q(t2)). t1

(124)

The endpoint cost varies as

δΦ = Φ (q(t))η(t ). q 2 2

(125)

Therefore the total terminal contribution is

[p(t2) + Φq (q (t2))]η(t2).

(126)

Stationarity for arbitrary η(t₂) gives

|---------------------| p(t2) + Φq(q(t2)) = 0. | -----------------------

(127)

This is the mechanics form of the natural boundary-condition logic developed in CV05: generalized momentum is literally the coefficient multiplying the free endpoint displacement in the first variation.

3 What these exercises add to CV14

The examples above expose several structural facts that are easy to miss if Hamilton’s principle is treated only as a formal recipe.

A stationary path can sometimes be proved globally minimizing. The free-particle action is a convex quadratic functional, so the exact difference formula proves global minimality.
Stationarity can survive after minimality is lost. The harmonic oscillator remains governed by the same Euler–Lagrange equation beyond its first conjugate time.
Symmetry becomes a first integral. Cyclic coordinates produce conserved generalized momenta.
Autonomy produces energy conservation. The identity _L = −L_t makes this precise.
Generalized coordinates change the formulas, not the variational logic. Coordinate-dependent inertia produces extra terms automatically.
Equivalent Lagrangians need not have identical canonical momenta. Total time derivatives preserve trajectories but shift p.
Momentum is a boundary coefficient. Free endpoint positions immediately produce natural momentum conditions.

These points are standard in analytical mechanics treatments [1, 2, 4]. They also prepare the next main lesson, CV15, where constraints and generalized coordinates are developed systematically.

Summary

Hamilton’s principle starts with

∫ ⋅ S[q] = L (q, q,t) dt

(128)

and demands stationarity under admissible path variations. The resulting Euler–Lagrange equation

|--------------| |d-L ⋅− Lq = 0| -dt-q-----------

(129)

contains Newtonian dynamics, but its variational origin also reveals more structure: generalized momentum, cyclic-coordinate conservation, energy conservation, natural endpoint conditions, and the distinction between stationary and minimizing trajectories.

The exercises show that the same framework handles straight-line free motion, projectile trajectories, oscillations, pendulum dynamics, central forces, coordinate-dependent kinetic energy, equivalent Lagrangians, and terminal boundary conditions. The calculations therefore provide a practical bridge from the formal calculus of variations to classical analytical mechanics.

References

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

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

[3] L. D. Landau and E. M. Lifshitz, Mechanics, 3rd ed., Butterworth-Heinemann, 1976.

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

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

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Calculus of Variations: Hamilton's Principle and Classical Mechanics - Exercises