Calculus of Variations: Free-Endpoint Problems and Worked Solutions

CV05 showed that the Euler–Lagrange equation governs the interior of a stationary curve while the endpoint freedom determines additional boundary conditions. This companion entry turns that principle into a problem-solving routine. The central fixed-x boundary term is

(1)

If an endpoint value is prescribed, the corresponding variation vanishes. If the endpoint value is free, the variation is arbitrary and the coefficient of that variation must vanish. This is the source of the natural boundary condition

(2)

When an endpoint cost is present, the same reasoning modifies the natural condition rather than replacing it. When the endpoint abscissa itself moves, the full transversality boundary form must be used. These constructions are standard in the classical calculus of variations and in analytical mechanics [1, 2, 3, 4].

1 How to use this set

Attempt every exercise before reading the solutions. For each problem use the same sequence:

1. identify what is fixed and what is free at each endpoint;
2. write the admissible variation conditions;
3. derive the Euler–Lagrange equation in the interior;
4. retain the boundary term instead of discarding it automatically;
5. impose the natural or transversality condition implied by the endpoint freedom;
6. only then solve for the integration constants; and
7. distinguish stationarity from any stronger minimum claim.

2 Exercises

Exercise 1: classify the endpoint data

For

(3)

assume the independent-variable endpoints a and b are fixed. For each case below, state the endpoint conditions on the variation η and the resulting natural boundary condition, if any.

1. y(a) and y(b) are both prescribed.
2. y(a) is prescribed and y(b) is free.
3. y(a) is free and y(b) is prescribed.
4. both y(a) and y(b) are free.

Explain why a free endpoint does not mean “there is no boundary condition.”

Exercise 2: the simplest one-free-end problem

Let

(4)

with

(5)

while y(L) is free.

1. Derive the Euler–Lagrange equation.
2. Derive the natural boundary condition at x = L.
3. Solve for the stationary function.
4. Prove directly that the stationary function is the global minimizer.

Exercise 3: a uniformly forced strip with a free right end

Consider

(6)

with

(7)

and y(L) free.

1. Derive the interior Euler–Lagrange equation.
2. Derive the natural condition at x = L.
3. Solve for y(x).
4. Find y(L) and verify that the terminal slope is zero.
5. Interpret Ty′ as the generalized boundary traction associated with the slope variable.

Exercise 4: a terminal spring and a Robin-type condition

Let

(8)

with y(0) = 0 and y(L) free.

1. Compute the first variation, including the variation of the terminal spring energy.
2. Derive the interior equation.
3. Show that the terminal condition is

   (9)

4. Solve the complete boundary-value problem.
5. Examine the limits k → 0 and k →∞ and interpret both.

Exercise 5: a prescribed endpoint load

Consider

(10)

with y(0) = 0 and y(L) free.

1. Derive the natural terminal condition.
2. Solve for y(x).
3. Show that the boundary condition can be written

   (11)

4. Explain why the free-end condition is not y′(L) = 0 in this problem.

Exercise 6: both endpoint values free

Let

(12)

and suppose both y(0) and y(L) are free.

1. Derive the Euler–Lagrange equation.
2. Derive both natural boundary conditions.
3. Solve the resulting boundary-value problem.
4. Prove directly that the stationary solution is the unique global minimizer.

Exercise 7: free terminal position in mechanics

A particle moves vertically in a uniform gravitational field. Consider the action

(13)

where t_f is fixed, q(0) = q₀ is prescribed, and q(t_f) is free.

1. Derive the equation of motion.
2. Derive the natural condition at t = t_f.
3. Solve the trajectory completely.
4. Explain the physical meaning of the natural condition in terms of canonical momentum.

Exercise 8: a conceptual free-terminal-time calculation

For a mechanical action

(14)

suppose the initial endpoint is fixed while the terminal point may move. CV05 showed that the terminal boundary form can be written

(15)

where

(16)

1. If t_f is fixed and q_f is free, what terminal condition follows?
2. If q_f is fixed and t_f is free, what terminal condition follows?
3. If both q_f and t_f are free independently, what two conditions follow?
4. Explain why these are necessary stationarity conditions rather than minimum tests.

The appearance of p and H in the endpoint variation is the classical mechanics form of transversality and is closely related to terminal conditions in optimal control [4, 5, 6].

3 Complete worked solutions

Solution 1: classify the endpoint data

For fixed a and b, integration by parts gives

(17)

On a stationary curve the interior Euler–Lagrange expression vanishes, so the endpoint logic is controlled entirely by

(18)

(a) Both endpoint values prescribed.

(19)

Hence the boundary term vanishes automatically. There are no additional natural conditions.

(b) Left value prescribed, right value free.

(20)

Therefore

(21)

(c) Left value free, right value prescribed.

(22)

so

(23)

(d) Both values free.

Both endpoint variations are independent and arbitrary, so

(24)

The key conceptual point is that freedom makes the endpoint variation arbitrary. An arbitrary variation is more restrictive on the coefficient, not less restrictive. Thus a free endpoint usually creates a natural boundary condition.

Solution 2: the simplest one-free-end problem

The integrand is

(25)

Hence

(26)

The Euler–Lagrange equation is

(27)

Since T is constant,

(28)

At the free endpoint x = L,

(29)

or simply

(30)

The general solution is

(31)

The natural condition gives C₁ = 0, and y(0) = Y ₀ gives C₂ = Y ₀. Therefore

(32)

To classify the solution, note that

(33)

For y_∗ = Y ₀, y_∗′ = 0 and therefore

(34)

Thus

(35)

for every admissible y. The stationary function is therefore a global minimizer, not merely a stationary candidate.

Solution 3: a uniformly forced strip with a free right end

Here

(36)

Thus

(37)

The Euler–Lagrange equation is

(38)

so

(39)

Because y(L) is free while L is fixed,

(40)

Integrating twice,

(41)

(42)

The fixed left endpoint gives C₂ = 0. The free-end condition gives

(43)

hence

(44)

Therefore

(45)

At the free end,

(46)

Also

(47)

so indeed

(48)

The quantity F_y′ = Ty′ is the quantity conjugate to the endpoint value y. In this simple one-dimensional elastic model it plays the role of a boundary traction. With no externally imposed endpoint load, stationarity sets that traction to zero.

Solution 4: a terminal spring and a Robin-type condition

Write

(49)

with

(50)

The variation of the integral is

(51)

The terminal spring contributes

(52)

Because y(0) is fixed, η(0) = 0. Therefore

(53)

Interior stationarity gives

(54)

Since y(L) is free, η(L) is arbitrary, so

(55)

The interior solution is

(56)

From y(0) = 0, D = 0. Hence y = Cx and the terminal condition becomes

(57)

Therefore

(58)

so

(59)

The stationary profile is

(60)

The endpoint value is

(61)

If k → 0, then C → 0 and the endpoint behaves like an unloaded free end. If k →∞, then

(62)

so a very stiff terminal spring approaches a prescribed endpoint value. This is a useful bridge between natural and essential boundary data.

Solution 5: a prescribed endpoint load

The integral part gives

(63)

The terminal load term contributes

(64)

Thus

(65)

The interior equation is

(66)

and the free terminal value gives

(67)

Since y = Cx + D and y(0) = 0, D = 0. The terminal condition gives

(68)

so

(69)

The free endpoint does not imply y′(L) = 0 because an external endpoint term is present. The natural condition expresses a balance between the internal conjugate quantity Ty′ and the applied terminal load P.

Solution 6: both endpoint values free

The integrand is

(70)

Therefore

(71)

The Euler–Lagrange equation is

(72)

or

(73)

Because both endpoint values are free,

(74)

The general solution is

(75)

Differentiate:

(76)

At x = 0,

(77)

so B = 0. Then

(78)

Since λ > 0 and L > 0, sinh(λL) > 0, hence A = 0. Therefore

(79)

Moreover,

(80)

with equality only if y′ = 0 and y = 0 throughout the interval. Thus y = 0 is the unique global minimizer.

Solution 7: free terminal position in mechanics

Here the independent variable is time and

(81)

The Euler–Lagrange equation is

(82)

Since

(83)

we obtain

(84)

or

(85)

Because the terminal time t_f is fixed but q(t_f) is free, the natural condition is

(86)

Thus

(87)

or

(88)

Integrating the equation of motion,

(89)

The natural condition gives

(90)

so

(91)

Integrating again,

(92)

The prescribed initial value gives C₂ = q₀. Therefore

(93)

The final position is

(94)

The canonical momentum is

(95)

Thus a free terminal position at fixed terminal time requires

(96)

For this example, the stationary trajectory reaches its turning point exactly at the terminal time.

Solution 8: a conceptual free-terminal-time calculation

The terminal boundary form is

(97)

(a) Fixed terminal time, free terminal position.

Here

(98)

while δq_f is arbitrary. Therefore

(99)

(b) Fixed terminal position, free terminal time.

Here

(100)

while δt_f is arbitrary. Therefore

(101)

(c) Both terminal position and terminal time free independently.

The variations δq_f and δt_f are independent, so both coefficients must vanish:

(102)

(d) Necessary versus sufficient.

These conditions come from requiring the first-order variation to vanish. They therefore identify stationary candidates. They do not determine whether the candidate is a minimum, maximum, or saddle. Classification requires additional information such as convexity or second-variation tests.

4 Common mistakes

• Mistake: setting η = 0 at every endpoint automatically. This silently converts a free-endpoint problem back into a fixed-endpoint one.
• Mistake: imposing F_y′ = 0 at an endpoint whose value is prescribed. The variation already vanishes there, so no new natural condition follows.
• Mistake: assuming every free endpoint gives y′ = 0. The actual condition is F_y′ = 0, or a modified balance when endpoint costs or loads are present.
• Mistake: forgetting the variation of an endpoint energy such as Φ(y(L)).
• Mistake: treating the natural condition as a replacement for the Euler–Lagrange equation. A stationary solution normally must satisfy both the interior equation and the relevant endpoint condition.
• Mistake: concluding “minimum” solely from Euler–Lagrange and natural boundary conditions. They are first-order necessary conditions.

5 Summary of skills practiced

After completing this set, the reader should be able to

• translate endpoint freedom into conditions on the variation η;
• derive natural boundary conditions from the retained integration-by-parts term;
• solve one-free-end and two-free-end boundary-value problems;
• include endpoint loads and terminal spring energies consistently;
• recognize Robin-type conditions as variational boundary balances;
• interpret F_y′ as an endpoint-conjugate quantity such as traction or canonical momentum;
• apply the fixed-terminal-time free-position condition p_f = 0 in mechanics;
• interpret the free-terminal-time condition H_f = 0 when the final coordinate is fixed; and
• keep stationarity separate from minimum classification.

CV05E2 will use the same endpoint boundary form when the terminal point is not fully free but constrained to lie on a prescribed curve. That restriction couples the allowed endpoint variations and produces the classical transversality relation.

References

References