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[parent] Minimum-Effort Interception of a Moving Target (Example)

Minimum-Effort Interception of a Moving Target: Transversality as a Bridge to Optimal Control

Purpose of this example

CV05 introduced transversality as the endpoint condition that appears when a terminal point is not fixed, but is constrained to lie on a curve. The phrase “endpoint constrained to a curve” can sound geometrically abstract. This example gives it an immediate physical interpretation: a vehicle must intercept a target whose position changes with time.

The target trajectory itself is the terminal constraint curve. The interception point is not known beforehand. The optimizer is allowed to choose when and where on that curve the interception occurs. The Euler–Lagrange equation determines the stationary motion in the interior; the transversality condition determines how that motion must meet the moving terminal constraint.

The example is deliberately based on the first-order kinematic model

′ x = u,
(1)

so that the transversality idea is not hidden by higher-order dynamics. The control u is a directly commanded velocity. Later optimal-control models may replace this by acceleration, thrust, steering angle, or another physical input.

1 Problem statement: intercept a moving target

A one-dimensional vehicle starts at

x(0) = 0.
(2)

A target begins a distance d > 0 ahead and moves away at constant speed V > 0:

x (t) = d + Vt. target
(3)

The interception time T is not prescribed. At interception, the vehicle must satisfy

x(T ) = d + V T.
(4)

Thus the terminal point (T,x(T)) may slide anywhere along the target line in the (t,x) plane.

We minimize the quadratic control-effort functional

∫ 1- T 2 J [u ] = 2 u (t)dt. 0
(5)

Using u = x′, this becomes the variational problem

|--------∫--T---------| J [x ] = 1 x′2(t)dt | | 2 0 | ----------------------
(6)

subject to Eqs. (2) and (4), with T free.

The phrase minimum effort here means minimum quadratic control cost. It is not, in general, literal fuel consumption or mechanical work. The example is intended to expose the variational endpoint structure as cleanly as possible.[3, 4]

PIC

Figure 1. The terminal point is not a prescribed point. It may slide along the moving target trajectory x = d + V t. For each trial interception time, the stationary interior path is a straight line. Transversality selects the particular point on the target curve at which the optimum occurs.

2 Why this is a transversality problem

In the generic notation of CV05, the independent variable was called x and the dependent function was y(x). Here it is more natural to use time t as the independent variable and position x(t) as the dependent function. The translation is therefore

xCV05 −→ t, yCV05 − → x(t).
(7)

The terminal point is

B = (T, x(T)).
(8)

Because T is free, the endpoint may move horizontally in the (t,x) plane. Because the vehicle must remain on the target trajectory at interception, the vertical terminal displacement is not independent of that horizontal motion.

For the target curve

x = ψ(t) = d + Vt,
(9)

an allowed endpoint displacement obeys the linearized constraint

|------------------------| |δxT = ψ ′(T )δT = V δT. | -------------------------
(10)

Equation (10) says that the endpoint may move, but only tangent to the target curve. This is exactly the endpoint geometry for which CV05 derived a transversality condition.

3 Deriving the moving-endpoint variation for this problem

Because the movement of the endpoint is central to this example, we derive the terminal variation rather than simply quoting the final formula.

Introduce a one-parameter family of nearby curves

x𝜖(t) = x (t) + 𝜖η(t),
(11)

and allow the terminal time to vary as

T𝜖 = T + 𝜖δT.
(12)

The perturbed functional is

∫ T𝜖 J(𝜖) = F(t,x ,x′) dt, 0 𝜖 𝜖
(13)

where, for the present problem,

F (t,x, x′) = 1-x′2. 2
(14)

The first variation is

| dJ | δJ = ---|| . d𝜖 𝜖=0
(15)

The upper integration limit depends on 𝜖, so the Leibniz rule gives

∫ T δJ = (Fxη + Fx ′η′)dt + F (T)δT. 0
(16)

The final term has a simple geometric interpretation. Moving the upper limit from T to T + δT adds, to first order, a thin strip with width δT and height F(T), hence the contribution F(T)δT.

Integrate the second term in Eq. (16) by parts:

∫ 0T F x′η′dt = [Fx′η] 0T −∫ 0T -d dtFx′η dt. (17)

Therefore

∫ ( ) T -d T δJ = Fx − dtFx ′ ηdt + [Fx ′η ]0 + F (T)δT. 0
(18)

The initial point is fixed, so

η(0) = 0.
(19)

For a stationary curve, the interior Euler–Lagrange equation

d Fx − --Fx′ = 0 dt
(20)

eliminates the integral term. The terminal contribution becomes

δJT = Fx′(T)η(T ) + F(T )δT.
(21)

4 The distinction between η(T) and the actual endpoint motion

Equation (21) contains η(T), but when the terminal time moves, η(T) is not the same quantity as the actual vertical endpoint displacement δxT .

The perturbed endpoint position is

x𝜖(T𝜖) = x𝜖(T + 𝜖δT ).
(22)

Substituting Eq. (11),

x𝜖(T𝜖) = x(T + 𝜖δT ) + 𝜖η(T + 𝜖δT ).
(23)

Taylor-expand to first order in 𝜖:

x(T + 𝜖δT) = x(T) + 𝜖x′(T)δT + O(𝜖2), (24)
𝜖η(T + 𝜖δT) = 𝜖η(T) + O(𝜖2). (25)

Hence

x 𝜖(T𝜖) = x(T ) + 𝜖[η (T ) + x′(T )δT ] + O (𝜖2).
(26)

The actual first-order terminal displacement is therefore

|----------------------| δxT = η(T ) + x ′(T )δT. ------------------------
(27)

Equivalently,

|----------------------| η (T) = δxT − x ′(T )δT.| ------------------------
(28)

This is the same moving-endpoint linearization developed in CV05. One term comes from deforming the curve at fixed t = T; the other comes from sliding the endpoint along the original curve.

Substitute Eq. (28) into Eq. (21):

δJT = Fx′ ′ (δxT − x δT) + FδT (29)
= Fx′δxT + (F − x ′Fx ′) δT. (30)

Thus the universal terminal boundary form in the (t,x) plane is

|--------------------------------| |δJT = Fx′ δxT + (F − x′Fx′)δT. | ---------------------------------
(31)

5 Apply the target-curve constraint

The endpoint must remain on the moving target line. Therefore Eq. (10) gives

δxT = V δT.
(32)

Substitute into Eq. (31):

δJT = Fx′V δT + (F − x′Fx′) δT (33)
= ′ ′ [F + (V − x)Fx ] δT. (34)

The interception point is free to move along the target trajectory, so δT is arbitrary. Stationarity therefore requires

|---------------------------------| F + (V − x′)Fx′ = 0 at t = T. | -----------------------------------
(35)

This is the transversality condition for this moving-target problem.

The key contrast with a natural boundary condition is worth emphasizing. A natural boundary condition appears when a terminal coordinate is freely variable by itself. Here the terminal variations are linked:

δxT = V δT.
(36)

The endpoint has one degree of freedom, but that freedom lies tangent to the target curve. Transversality is the stationarity condition associated with that constrained endpoint motion.

6 Euler–Lagrange equation in the interior

For

1 ′2 F = -x , 2
(37)

we have

Fx = 0, Fx ′ = x ′.
(38)

The Euler–Lagrange equation is therefore

-d ′ 0 − dt (x) = 0,
(39)

or

|-′′-----| -x--=-0.-|
(40)

Integrating twice,

x(t) = C0 + C1t.
(41)

The fixed initial condition x(0) = 0 gives C0 = 0, so

|----------------------------------| |x(t) = C1t, u(t) = x′(t) = C1. | -----------------------------------
(42)

Thus the stationary interior trajectory uses a constant velocity command. The remaining question is: which constant velocity, and which interception point on the target curve? That is precisely what the terminal constraint and transversality condition determine.

7 Use transversality to determine the optimal control

Substitute

1- ′2 ′ F = 2 x , Fx′ = x
(43)

into Eq. (35):

1 ′2 ′ ′ -x + (V − x )x = 0. 2
(44)

Expand:

1 --x′2 + V x′ − x′2 = 0, 2
(45)

so

V x′ − 1-x′2 = 0. 2
(46)

Factor:

( ) ′ 1- ′ x V − 2 x = 0.
(47)

The root x′ = 0 cannot intercept a target that begins at d > 0. The nontrivial stationary solution is therefore

|-′----∗-------| -x-=--u--=-2V.-|
(48)

The terminal constraint now determines T:

2V T = d + V T.
(49)

Hence

|--------| T ∗ = d-.| ------V---
(50)

The interception position is

|------------| |x(T ∗) = 2d.| -------------
(51)

The target travels a distance d during the optimal interception interval, so it moves from x = d to x = 2d. The vehicle travels twice that distance from the origin, at twice the target speed.

8 What does the solution mean physically?

The result

u∗ = 2V, T ∗ = d- V
(52)

has a useful physical interpretation.

If the vehicle tries to intercept too early, the target has not moved very far, but the vehicle must cover the initial gap d in a very short time. The required command u is large, and the quadratic penalty u2 makes that expensive.

If the vehicle waits too long, the required excess speed above the target speed is smaller, but the target continues to move away and the control must be applied over a longer interval. The accumulated cost again becomes large.

The optimum balances these two effects. At T = d∕V , the target has moved exactly one initial-gap distance d. The vehicle has traveled 2d, and the relative closing speed is

u∗ − V = V.
(53)

Thus the time needed to close the original gap is

--d----= d-= T∗. u∗ − V V
(54)

The transversality condition has therefore selected both a terminal point and a control level with a simple physical meaning.

PIC

Figure 2. The stationary interior solution has constant control because x′′ = 0. Transversality selects the particular level u∗ = 2V . The target moves at speed V , so the optimal vehicle command is twice the target speed.

9 An independent check: optimize directly over interception time

There is a useful way to verify the transversality result without using the moving-endpoint formula.

Suppose temporarily that a trial interception time T has been chosen. The terminal point is then fixed at

x(T ) = d + V T.
(55)

For this fixed T, the Euler–Lagrange equation gives a constant-speed path from 0 to d + V T:

d + V T d uT = --------= V + --. T T
(56)

The corresponding cost is

J(T) = 1 -- 2uT 2T (57)
= 1- 2( ) d- V + T2T (58)
= 2 d-- 2T + dV + 2 V--T- 2 . (59)

Differentiate:

dJ d2 V 2 ---= − --2-+ ---. dT 2T 2
(60)

Stationarity gives

2 d-- = V 2. T 2
(61)

Because T > 0, d > 0, and V > 0,

T = d-, V
(62)

exactly as obtained from transversality.

Moreover,

d2J d2 ---2 = -3-> 0, dT T
(63)

so this stationary value is a strict minimum with respect to the interception time.

At the optimum, the two T-dependent contributions in Eq. (59) are equal:

2 2 ∗ -d--= V--T--= dV-. 2T ∗ 2 2
(64)

This equality gives another interpretation of the optimum: the cost associated with closing the initial spatial gap and the cost associated with the target’s continued motion are balanced.

PIC

Figure 3. Normalized cost versus normalized interception time. Very early interception is expensive because the required speed is large; very late interception is expensive because the control is maintained while the target continues to move away. The minimum occurs at V T∕d = 1.

10 Minimum cost

Substitute Eqs. (48) and (50) into the objective:

J∗ = 1 -- 2 ∫ 0d∕V (2V )2 dt (65)
= 1- 2(4V 2)d- V. (66)

Therefore

|----------| -J∗-=-2dV.-|
(67)

The units are those of the mathematical quadratic control cost. Since u is a velocity command here,

2 [J ] = length--. time
(68)

This should not be interpreted automatically as mechanical energy.

11 Numerical example

Take

d = 100 m, V = 10 m ∕s.
(69)

The transversality solution gives

∗ u = 2V = 20 m ∕s,
(70)

and

∗ 100 T = ----= 10 s. 10
(71)

The target begins at 100 m and moves another 100 m during those ten seconds, so the interception occurs at

x(T ∗) = 200 m.
(72)

The minimum quadratic control cost is

J∗ = 2 (100 )(10 ) = 2000 m2 ∕s.
(73)

The vehicle’s relative closing speed is

20 − 10 = 10 m ∕s,
(74)

so closing the initial 100 m gap takes exactly 10 s.

12 Geometric meaning of transversality

Define the momentum-like endpoint coefficient

p = Fx ′
(75)

and the Hamiltonian-like quantity

ℋ = x′p − F.
(76)

Then the terminal boundary form, Eq. (31), becomes

|--------------------| |δJT = p δxT − ℋ δT. | ----------------------
(77)

The terminal target curve has tangent displacement

(δT, δxT) = δT (1,V ).
(78)

Equation (77) can therefore be written as the dot product

δJT = (− ℋ, p) ⋅ (δT,δxT ).
(79)

Stationarity requires this dot product to vanish for every allowed tangent motion. Hence

|------------------| |(− ℋ, p) ⊥ (1,V ). | -------------------
(80)

The boundary covector is normal to the terminal constraint curve. This is the geometric content of the word transversality.

For the present integrand,

1 1 p = x ′ = u, ℋ = u2 − -u2 = -u2. 2 2
(81)

Thus Eq. (80) gives

− 1u2 + uV = 0, 2
(82)

which again yields the nontrivial solution u = 2V .

PIC

Figure 4. At the selected terminal point, allowed endpoint motion is tangent to the target curve. The variational boundary covector (−ℋ,p) must annihilate that tangent motion and is therefore normal to the target curve. This orthogonality in endpoint-variation space is the geometric meaning of transversality.

13 The same result in optimal-control language

Now rewrite the problem exactly as an elementary optimal-control problem:

∫ ′ T1- 2 x = u, J = 2 u dt. 0
(83)

Use the minimization-convention control hamiltonian

1 Hc = -u2 + λu. 2
(84)

Stationarity with respect to u gives

∂Hc- = u + λ = 0, ∂u
(85)

so

|--------| -u-=-−-λ.-
(86)

The costate equation is

∂H λ′ = − ---c= 0, ∂x
(87)

so the costate is constant, and therefore the optimal control is constant. This is the same conclusion obtained from the Euler–Lagrange equation x′′ = 0.

The terminal condition is not a fixed terminal state. It is the terminal manifold

g(T, x(T )) = x (T ) − d − V T = 0.
(88)

Introduce a terminal multiplier ν. The standard transversality conditions for this equality constraint, with no terminal cost, are

λ(T ) = νgx,
(89)

and

Hc (T) + νgT = 0.
(90)

For Eq. (88),

gx = 1, gT = − V.
(91)

Therefore

λ(T ) = ν,
(92)

and

Hc (T) − νV = 0.
(93)

From Eq. (86), ν = λ = −u. On the stationary control,

1-2 2 1- 2 Hc = 2u − u = − 2 u .
(94)

Thus the terminal condition becomes

1 2 − -u + uV = 0, 2
(95)

and again

u ∗ = 2V.
(96)

The same terminal geometry has therefore appeared in three equivalent forms:

|------------------------| |CV endpoint condition | | F + (V − x′)Fx ′ = 0 | --------------------------
(97)

|----------------| (−-ℋ,-p)-⊥-(1,V-)-
(98)

and

|----------------------------------| |λ(T ) = νgx, Hc (T) + νgT = 0.| -----------------------------------
(99)

The first is classical calculus of variations, the second is its geometric boundary interpretation, and the third is the terminal transversality language of optimal control.[3, 4]

14 Natural boundary condition versus transversality

CV05E2 considered a free terminal state coordinate. This article considers a terminal point constrained to a curve. The distinction can be summarized as follows.

Endpoint type

Allowed variation

Boundary consequence

Fixed terminal coordinate

Variation of that coordinate is zero

No new boundary equation from that coordinate

Free terminal coordinate

Variation is arbitrary and independent

Its conjugate boundary coefficient must vanish

Endpoint constrained to a curve

Endpoint variations are nonzero but linked by the curve tangent

A linear combination of boundary coefficients must vanish: transversality

In the present example,

δxT = VδT
(100)

is the crucial relation. Neither δxT nor δT is forced to zero, but they cannot vary independently. That is why the terminal condition is not simply Fx′ = 0 or F − x′Fx′ = 0. Instead the two coefficients combine according to the tangent direction of the target curve.

15 Generalization to a nonlinear moving target

Suppose the target trajectory is a smooth nonlinear function

xtarget(t) = ψ(t).
(101)

Then an allowed terminal displacement satisfies

δxT = ψ′(T)δT.
(102)

The transversality condition becomes

|----------------------------| |F + (ψ ′(T ) − x ′(T ))F ′ = 0. ----------------------x-------
(103)

For F = 1 2x′2, the nontrivial terminal condition reduces to

|----------------| | ′ ′ | -x-(T)-=-2ψ-(T-).
(104)

The interior Euler–Lagrange equation still requires x′ to be constant in this simple single-integrator model, so the interception time must satisfy both

x (T ) = ψ (T )
(105)

and

′ ′ x = 2 ψ (T ).
(106)

For a nonlinear target path, these equations may need to be solved numerically. This is already very close in spirit to practical trajectory optimization.

16 What should be carried forward into optimal control?

This example provides several ideas that recur in more advanced trajectory optimization:

  1. The terminal state need not be a prescribed point. It may lie on a terminal curve, surface, or higher-dimensional manifold.
  2. The Euler–Lagrange equations govern the stationary trajectory in the interior, but they do not by themselves determine which terminal point on the constraint should be selected.
  3. Allowed endpoint variations are tangent to the terminal constraint.
  4. The variational boundary covector must annihilate every allowed tangent displacement; geometrically, it is normal to the terminal constraint.
  5. In optimal control, the same normality condition is expressed through terminal costates and multipliers.
  6. A transversality condition is a necessary stationarity condition. It does not by itself prove that the stationary trajectory is a minimum.

For the moving-target problem, the conceptual chain is

|------------------------------------------------------------------------------| | moving target −→ terminal curve −→ linked endpoint variations | | −→ transversality − → terminal manifold in optimal control | -------------------------------------------------------------------------------|
(107)

Summary

A kinematic vehicle obeying

x′ = u

must intercept a target

xtarget(t) = d + V t

while minimizing

∫ 1- T 2 J = 2 u dt. 0

The final time is free, and the final point must lie on the target trajectory. The moving-endpoint first variation produces

δJT = Fx′δxT + (F − x′Fx ′) δT.

Because the target curve imposes

δxT = V δT,

stationarity requires

F + (V − x ′)Fx′ = 0.

For

1 F = -x′2, 2

Euler–Lagrange gives constant x′, while transversality selects

|-∗------| u--=--2V.-

The terminal constraint then gives

|----------------------------------------| | d | |T ∗ = --, x (T ∗) = 2d, J∗ = 2dV. | -------V---------------------------------

The example shows concretely what it means for an endpoint to be “constrained to a curve.” The terminal point may move, but only tangent to the target trajectory. Transversality supplies the boundary equation associated with that constrained motion, and the same structure becomes a terminal-manifold condition in optimal control.

References

References

[1]   I. M. Gelfand and S. V. Fomin, Calculus of Variations, Prentice-Hall, 1963; reprinted by Dover Publications.

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

[3]   D. Liberzon, Calculus of Variations and Optimal Control Theory: A Concise Introduction, Princeton University Press, 2012.

[4]   D. E. Kirk, Optimal Control Theory: An Introduction, Prentice-Hall, 1970.


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Other names:  CV05E3
Keywords:  calculus of variations, transversality, moving endpoint, terminal constraint, moving target, interception, minimum control effort, optimal control, terminal manifold, costate, Hamiltonian

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Cross-references: relation, type, manifold, hamiltonian, dot product, energy, units, boundary, formula, displacement, function, work, speed, acceleration, velocity, kinematic, motion, position, CV05

This is version 2 of Minimum-Effort Interception of a Moving Target, born on 2026-09-27, modified 2026-09-27.
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 02.30.Sa (Functional analysis)
 45.20.Jj (Lagrangian and Hamiltonian mechanics)
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