Bolza, Mayer, and Lagrange Forms, Minimum-Time Conventions, and Singular Arcs

Bolza, Mayer, and Lagrange Forms¶

The most general standard cost functional is the Bolza form:

\boxed{ J = \Phi\!\left( \boldsymbol{x}(t_0),t_0, \boldsymbol{x}(t_f),t_f \right) + \int_{t_0}^{t_f} L(\boldsymbol{x},\boldsymbol{u},t)\,\mathrm{d} t. }

(17)

It contains both an endpoint cost and a running cost.

Mayer form¶

The Mayer form contains only an endpoint cost:

\boxed{ J = \Phi\!\left( \boldsymbol{x}(t_0),t_0, \boldsymbol{x}(t_f),t_f \right). }

(18)

Lagrange form¶

The Lagrange form contains only a running cost:

\boxed{ J = \int_{t_0}^{t_f} L(\boldsymbol{x},\boldsymbol{u},t)\,\mathrm{d} t. }

(19)

Relationship among the forms¶

Form	Endpoint term	Integral term
Mayer	Present	Absent
Lagrange	Absent	Present
Bolza	Present	Present

Converting Lagrange Form to Mayer Form¶

Consider a Lagrange-form problem:

J = \int_{t_0}^{t_f} L(\boldsymbol{x},\boldsymbol{u},t)\,\mathrm{d} t,

(20)

subject to

\dot{\boldsymbol{x}} = \boldsymbol{f}(\boldsymbol{x},\boldsymbol{u},t).

(21)

Introduce an additional state:

x_{n+1}(t) = \int_{t_0}^{t} L(\boldsymbol{x}(\tau),\boldsymbol{u}(\tau),\tau)\,\mathrm{d}\tau.

(22)

Then

\dot{x}_{n+1} = L(\boldsymbol{x},\boldsymbol{u},t),

(23)

with

x_{n+1}(t_0)=0.

(24)

At the final time,

x_{n+1}(t_f) = \int_{t_0}^{t_f} L(\boldsymbol{x},\boldsymbol{u},t)\,\mathrm{d} t = J.

(25)

Therefore, the original Lagrange problem is equivalent to the Mayer problem

\boxed{ \min x_{n+1}(t_f). }

(26)

Augmented-State Dynamics¶

Define the augmented state

\boldsymbol{y} = \begin{bmatrix} \boldsymbol{x}\\ x_{n+1} \end{bmatrix}.

(27)

Then

\dot{\boldsymbol{y}} = \begin{bmatrix} \boldsymbol{f}(\boldsymbol{x},\boldsymbol{u},t)\\ L(\boldsymbol{x},\boldsymbol{u},t) \end{bmatrix} = \boldsymbol{g}(\boldsymbol{y},\boldsymbol{u},t).

(28)

The augmented Mayer problem is

\min x_{n+1}(t_f),

(29)

subject to

\dot{\boldsymbol{y}}=\boldsymbol{g}(\boldsymbol{y},\boldsymbol{u},t).

(30)

Can Mayer Form Always Be Converted to Lagrange Form?¶

The reverse conversion is not always as direct.

A terminal cost

\Phi(\boldsymbol{x}(t_f),t_f)

(31)

may be nonlinear or may depend on endpoint variables in a way that is not naturally represented as a standard running cost.

In some cases, one can use identities such as

\Phi(\boldsymbol{x}(t_f)) - \Phi(\boldsymbol{x}(t_0)) = \int_{t_0}^{t_f} \frac{\,\mathrm{d}}{\,\mathrm{d} t} \Phi(\boldsymbol{x}(t))\,\mathrm{d} t,

(32)

but this generally introduces state derivatives and additional assumptions.

Therefore, the Lagrange-to-Mayer conversion is universal, whereas the reverse direction may be inconvenient or problem dependent.

Return to the Minimum-Time Double Integrator¶

The normalized dynamics are

\begin{aligned} \dot{x}_1&=x_2,\\ \dot{x}_2&=u, \end{aligned}

(33)

with

-1\leq u\leq 1.

(34)

Using the Mayer cost

J=t_f,

(35)

the Hamiltonian is

H = \lambda_1x_2+\lambda_2u.

(36)

Because the problem is autonomous,

H^*(t)=\text{constant}.

(37)

The free-final-time transversality condition yields

\boxed{ H^*(t)\equiv -1. }

(38)

Costate Equations¶

The costate equations are

\begin{aligned} \dot{\lambda}_1 &= -\frac{\partial H}{\partial x_1} = 0, \\ \dot{\lambda}_2 &= -\frac{\partial H}{\partial x_2} = -\lambda_1. \end{aligned}

(39)

Hence,

\lambda_1(t)=C_1,

(40)

and

\lambda_2(t) = -C_1t+C_2.

(41)

Thus, $\lambda_2$ is affine in time.

Careful Interpretation of the Minimum Principle¶

Pontryagin’s Minimum Principle states that

H\!\left( \boldsymbol{x}^*(t),\boldsymbol{u}^*(t),\boldsymbol{\lambda}^*(t),t \right) \leq H\!\left( \boldsymbol{x}^*(t),\boldsymbol{u}(t),\boldsymbol{\lambda}^*(t),t \right)

(42)

for every admissible control $\boldsymbol{u}(t)\in\mathcal{U}$ .

The state and costate are held fixed at their optimal values. Only the control is varied.

Therefore,

u^*(t) \in \operatorname*{arg\,min}_{u\in[-1,1]} \left[ \lambda_1^*(t)x_2^*(t) + \lambda_2^*(t)u \right].

(43)

Since the first term is independent of $u$ ,

u^*(t) \in \operatorname*{arg\,min}_{u\in[-1,1]} \lambda_2^*(t)u.

(44)

Thus,

\boxed{ u^*(t) = -\operatorname{sgn}\!\left(\lambda_2^*(t)\right) }

(45)

when $\lambda_2^*(t)\neq 0$ .

Possible Shapes of the Switching Function¶

The switching function is

\sigma(t) = \lambda_2(t) = -C_1t+C_2.

(46)

The possibilities are:

positive for the entire interval;
negative for the entire interval;
positive initially and negative later;
negative initially and positive later; or
constant.

The first four cases correspond to zero or one switch.

The constant case requires

C_1=0.

(47)

Then

\lambda_1(t)\equiv 0,

(48)

and

\lambda_2(t)\equiv C_2.

(49)

Can the Switching Function Be Constant?¶

The autonomous Mayer-form Hamiltonian must satisfy

H^*(t) = \lambda_1^*x_2^* + \lambda_2^*u^* = -1.

(50)

If $C_1=0$ , then

\lambda_1^*(t)=0,

(51)

\lambda_2^*(t)u^*(t) = -1.

(52)

A nonzero constant $\lambda_2$ can satisfy this identity only for a compatible constant bang control and a specific multiplier normalization. Such cases do not introduce a singular arc because the sign of $\lambda_2$ still determines the control.

The important problematic case is

\lambda_2(t)\equiv 0.

(53)

Then

H^*(t)=0,

(54)

which contradicts

H^*(t)\equiv -1.

(55)

Therefore,

\boxed{ \lambda_2(t) \text{ cannot be identically zero on a finite interval.} }

(56)

Definition of a Singular Arc¶

Let the switching function be

\sigma(t) = \frac{\partial H}{\partial u}.

(57)

A singular arc or singular interval is a nonzero-duration interval

, \qquad t_2>t_1,

(58)

on which

\sigma(t)\equiv 0.

(59)

On such an interval, the usual Hamiltonian minimization condition does not determine the control directly.

For the minimum-time double integrator,

\sigma(t)=\lambda_2(t).

(60)

Since $\lambda_2$ cannot vanish identically on a finite interval, the standard minimum-time double integrator has no singular arc.

Isolated Switching Time Versus Singular Arc¶

At a normal switching time $t_s$ ,

\sigma(t_s)=0,

(61)

but

\sigma(t)\neq 0

(62)

immediately before and after the switch.

A singular arc instead satisfies

\sigma(t)=0

(63)

throughout an interval.

Therefore,

\boxed{ \text{An isolated zero of the switching function is not a singular arc.} }

(64)

General Mechanism Producing Singular Arcs¶

Consider

H = L(\boldsymbol{x},t) + \boldsymbol{\lambda}^{\mathsf{T}}\boldsymbol{f}(\boldsymbol{x},\boldsymbol{u},t),

(65)

where $L$ does not depend explicitly on the control.

If the control-dependent contribution becomes insensitive to $u$ along an interval, then Hamiltonian minimization cannot determine the optimal control there.

For a control-affine system,

\boldsymbol{f}(\boldsymbol{x},\boldsymbol{u},t) = \boldsymbol{f}_0(\boldsymbol{x},t) + \boldsymbol{B}(\boldsymbol{x},t)\boldsymbol{u},

(66)

the Hamiltonian is

H = L(\boldsymbol{x},t) + \boldsymbol{\lambda}^{\mathsf{T}}\boldsymbol{f}_0 + \boldsymbol{\lambda}^{\mathsf{T}}\boldsymbol{B}\boldsymbol{u}.

(67)

The switching function is

\boldsymbol{\sigma}(t) = \frac{\partial H}{\partial \boldsymbol{u}} = \boldsymbol{B}^{\mathsf{T}}\boldsymbol{\lambda}.

(68)

A singular arc occurs when

\boldsymbol{\sigma}(t)\equiv\boldsymbol{0}

(69)

on a finite interval.

Determining a Singular Control¶

On a singular interval, one differentiates the switching function until the control appears explicitly:

\begin{aligned} \boldsymbol{\sigma}(t)&=\boldsymbol{0},\\ \dot{\boldsymbol{\sigma}}(t)&=\boldsymbol{0},\\ \ddot{\boldsymbol{\sigma}}(t)&=\boldsymbol{0},\\ &\vdots \end{aligned}

(70)

Suppose the control first appears in the $r$ th derivative:

\boldsymbol{\sigma}^{(r)} = \boldsymbol{a}(\boldsymbol{x},\boldsymbol{\lambda},t) + \boldsymbol{B}_s(\boldsymbol{x},\boldsymbol{\lambda},t)\boldsymbol{u}.

(71)

Then the candidate singular control satisfies

\boxed{ \boldsymbol{u}_s = - \boldsymbol{B}_s^{-1}\boldsymbol{a}, }

(72)

provided the required inverse exists.

This candidate must also satisfy:

the control bounds;
the state and costate equations;
endpoint and continuity conditions;
generalized Legendre–Clebsch conditions; and
global consistency with neighboring arcs.

Why Singular Arcs Are Numerically Difficult¶

Singular arcs are challenging because:

the switching function may be very small without being identically zero;
bang and singular arcs may meet at unknown times;
high-order derivatives may be required;
numerical differentiation amplifies error;
control bounds must still be enforced; and
the resulting boundary-value problem becomes highly sensitive.

A numerical method may incorrectly approximate a singular arc as:

rapid switching;
chattering;
a smoothed bang-bang transition; or
an interior control with no correct theoretical basis.

Examples Where Singular Arcs Arise¶

Singular arcs appear in many applications, including:

Goddard rocket ascent;
chemical reactors;
bioreactor feeding;
aerospace trajectory optimization;
energy management;
thermal processes; and
systems with one-sided actuation.

In the Goddard rocket problem, the optimal thrust structure can contain:

\text{maximum thrust} \;\longrightarrow\; \text{singular thrust} \;\longrightarrow\; \text{minimum thrust}.

(73)

The middle segment cannot be determined from the basic switching-function sign alone.

One-Sided Control and Singular Behavior¶

Many mechanical systems permit two-sided actuation:

-u_{\max}\leq u\leq u_{\max}.

(74)

Many chemical and process systems permit only one-sided actuation:

0\leq u\leq u_{\max}.

(75)

For example, material may be added to a reactor but not removed through the same input channel.

One-sided control often creates:

asymmetric bang arcs;
boundary arcs;
long zero-control intervals; and
singular arcs.

Common Errors¶

Treating the Mayer and Lagrange minimum-time Hamiltonians as numerically identical.
Forgetting that the two Hamiltonians differ by a constant.
Calling a general endpoint-plus-integral cost “Mayer form” instead of Bolza form.
Adding an accumulated-cost state but forgetting its initial condition.
Minimizing the Hamiltonian while allowing the state and costate to vary.
Calling every zero of the switching function a singular arc.
Assuming that $\partial H/\partial u=0$ determines the control on a singular interval.
Ignoring Hamiltonian transversality when testing whether a singular arc is possible.

Summary¶

The main conclusions are:

Minimum time can be represented in Mayer or Lagrange form.
The Mayer Hamiltonian is constant at -1 under the convention used here.
The equivalent Lagrange Hamiltonian is constant at zero.
Bolza form contains both endpoint and running costs.
Every Lagrange problem can be converted to Mayer form by adding one accumulated-cost state.
Pontryagin’s Minimum Principle varies only the control while holding the optimal state and costate fixed.
The minimum-time double-integrator switching function is affine in time.
An identically zero switching function is incompatible with the Hamiltonian condition in this problem.
A singular arc is a finite interval on which the switching function vanishes identically.
Singular controls require differentiated switching conditions and additional optimality tests.

Connection. Bang–bang reasoning ceases to determine the control when the switching function vanishes over an interval; this is the origin of singular control.