Minimum-Fuel Control & Bang-Off-Bang Structure

Minimum-fuel control replaces “finish as soon as possible” with “use as little accumulated control magnitude as possible.” The absolute value changes the Hamiltonian geometry: small switching-function values make zero control optimal, while large values select saturation. That produces bang-off-bang trajectories and also creates existence pathologies when final time is free.

Absolute-effort model and its physical meaning #

For a scalar actuator,

$$J[u]=\int_{t_0}^{t_f}\rho(t)|u(t)|dt, \qquad |u(t)|\le u_{\max},$$

with $\rho(t)>0$. If $u$ is mass-flow command and the propulsion model makes $\rho|u|$ equal to propellant mass rate, the criterion is literal fuel. If $u$ is force, voltage, or a nondimensional input, it is an $L^1$ effort proxy until a physical calibration proves otherwise.

For multiple independent channels, the separable criterion is

$$J[u]=\int_{t_0}^{t_f}\sum_{i=1}^m\rho_i|u_i|dt.$$

This differs from the rotationally invariant group cost

$$J[u]=\int_{t_0}^{t_f}\rho\lVert u\rVert_2dt.$$

The first encourages individual channels to be exactly zero; the second switches the entire vector between zero and a radial boundary direction.

Scalar bang-off-bang law #

For the control-affine system

$$\dot x=f_0(t,x)+g(t,x)u,$$

the normal maximum-convention Hamiltonian is

$$H=p^Tf_0+\sigma u-\rho|u|, \qquad \sigma=p^Tg.$$

At each time, maximize

$$h(u)=\sigma u-\rho|u|, \qquad |u|\le u_{\max}.$$

For $u\ge0$, $h=(\sigma-\rho)u$; for $u\le0$, $h=(\sigma+\rho)u$. Therefore

$$\boxed{ u^*=\begin{cases} +u_{\max}, & \sigma>\rho,\\ 0, & |\sigma|<\rho,\\ -u_{\max}, & \sigma<-\rho. \end{cases}}$$

At $\sigma=\rho$, every $u\in[0,u_{\max}]$ gives the same instantaneous Hamiltonian. At $\sigma=-\rho$, every $u\in[-u_{\max},0]$ does. These threshold sets require higher-order dynamics or endpoint conditions to determine a singular candidate.

For componentwise box constraints and separable cost,

$$\sigma_i=(G^Tp)_i,$$

and every channel obeys its own threshold:

$$u_i^*=\begin{cases} +\bar u_i,&\sigma_i>\rho_i,\\ 0,&|\sigma_i|<\rho_i,\\ -\bar u_i,&\sigma_i<-\rho_i. \end{cases}$$

Euclidean vector input #

Let $\lVert u\rVert_2\le u_{\max}$ and use running cost $\rho\lVert u\rVert_2$. With $q=G^Tp$, maximize

$$q^Tu-\rho\lVert u\rVert_2.$$

Cauchy–Schwarz gives $q^Tu\le\lVert q\rVert_2\lVert u\rVert_2$. Hence

$$\boxed{ u^*=\begin{cases} 0, & \lVert q\rVert_2<\rho,\\[4pt] u_{\max}\dfrac{q}{\lVert q\rVert_2}, & \lVert q\rVert_2>\rho. \end{cases}}$$

At $\lVert q\rVert_2=\rho$, any magnitude from zero to $u_{\max}$ along $q$ is an instantaneous maximizer. The direction comes from the dual vector $q$; the dead-zone test comes from its dual norm. For a general control norm, the corresponding dual norm appears.

Minimum fuel is not minimum time #

Adding a small time weight,

$$J=\int(\rho|u|+\alpha)dt, \qquad\alpha>0,$$

penalizes waiting and changes the thresholds and optimum. It is a legitimate new formulation, not a numerical trick that preserves the original minimum-fuel problem.

A lower bound from state balance #

For the double integrator

$$\dot x_1=x_2,qquad \dot x_2=u,$$

with terminal velocity $x_2(t_f)=0$,

$$\int_{t_0}^{t_f}u(t)dt=-x_2(t_0).$$

The triangle inequality gives

$$\boxed{ J=\int_{t_0}^{t_f}|u(t)|dt \ge\left|\int_{t_0}^{t_f}u(t)dt\right| =|x_2(t_0)|.}$$

Equality requires $u$ to have one sign almost everywhere, except on zero-control intervals. That may be incompatible with the position endpoint. The inequality is a lower bound, not proof that a feasible trajectory attains it.

Nonattainment with free final time #

Start from

$$x_1(0)=a>0,qquad x_2(0)=0,$$

and require $x(t_f)=(0,0)$ with $|u|\le1$. The velocity balance gives only $J\ge0$. No control with $J=0$ can move the state, because $J=0$ implies $u=0$ almost everywhere.

Nevertheless, for any $0<\delta\le1$ construct:

1. apply $u=-1$ for $\delta$ seconds, reaching velocity $-\delta$;
2. coast with $u=0$;
3. apply $u=+1$ for $\delta$ seconds, returning velocity to zero.

The acceleration and braking ramps each cover distance $\delta^2/2$, so the coasting duration is

$$T_c=\frac{a-\delta^2}{\delta}$$

when $a\ge\delta^2$. Total time and fuel are

$$t_f=2\delta+T_c =\frac{a}{\delta}+\delta,$$

$$J_\delta=\delta+\delta=2\delta.$$

As $\delta\downarrow0$,

$$J_\delta\to0, \qquad t_f\to\infty.$$

Thus

$$\inf J=0$$

but no finite-time admissible control attains it. PMP can characterize every finite candidate and still cannot manufacture a nonexistent minimizer.

Near-optimal controls when the infimum is not attained #

If $J_*=\inf_{u\in\mathcal U_{\mathrm{ad}}}J[u]$ is not attained, a control is $\mu$-optimal when

$$J[u_\mu]\le J_*+\mu, \qquad \mu>0.$$

For the construction above, choose $\delta\le\mu/2$ to obtain $J_\delta\le\mu$. The price is a duration approximately $a/\delta$, which becomes unbounded as tighter cost accuracy is demanded.

Numerical and implementation consequences #

• Smooth approximations such as $\sqrt{u^2+\varepsilon^2}$ remove the corner but change the exact dead-zone structure.
• Direct collocation may return a very long low-amplitude solution when the continuous problem has no minimizer; mesh convergence must include terminal time and cost, not only state residuals.
• Minimum-fuel switching is sensitive to model bias near $|\sigma|=\rho$. Hysteresis can prevent chatter but changes the switching law.
• Propellant systems often have minimum impulse bits, valve dwell times, and startup costs. These require hybrid or mixed-integer models; continuous $L^1$ control alone does not represent them.