Soft Lunar Landing on a Flat Moon

This section studies the classical 1964 George Meditch soft-landing problem. The objective is to land a lunar module with zero altitude and zero velocity while minimizing fuel consumption. The lecture emphasizes problem formulation and optimality conditions rather than carrying the algebra to completion.

Dynamics¶

\begin{aligned} \dot h &= v,\\ \dot v &= -g+u, \end{aligned}

(1)

where $h$ is altitude, $v$ vertical velocity, $u$ the thrust acceleration, and

0\le u\le 3.

(2)

Initial conditions:

h(0)=h_0,\qquad v(0)=v_0.

(3)

Terminal conditions:

h(t_f)=0,\qquad v(t_f)=0.

(4)

The terminal time is free.

Performance Index¶

Fuel consumption is approximated by

J=\int_0^{t_f}u(t)\,dt,

(5)

which is proportional to total thrust because $u\ge0$ .

Hamiltonian¶

H=1+\lambda_hv+\lambda_v(-g+u) =(1+\lambda_v)u+\lambda_hv-\lambda_v g.

(6)

Costate Equations¶

\begin{aligned} \dot\lambda_h&=0,\\ \dot\lambda_v&=-\lambda_h. \end{aligned}

(7)

Hence

\lambda_h=C_1,\qquad \lambda_v=-C_1t+C_2.

(8)

Pontryagin Minimum Principle¶

Because the Hamiltonian is linear in $u$ , the minimum principle is used directly:

u^*=\arg\min_{0\le u\le3}H.

(9)

Therefore

u^*= \begin{cases} 0,&1+\lambda_v>0,\\ 3,&1+\lambda_v<0. \end{cases}

(10)

A singular arc would require

1+\lambda_v\equiv0,

(11)

which cannot occur except in the degenerate case, so no singular arc exists.

Bang–Bang Structure¶

Since $\lambda_v$ is linear in time, $1+\lambda_v$ is also linear. Consequently, the optimal control has at most one switching time:

0\rightarrow3 \quad\text{or}\quad 3\rightarrow0,

(12)

or remains constant over the entire trajectory.

Free Final Time¶

Because $t_f$ is free and the Hamiltonian has no explicit time dependence,

H(t_f)=0,

(13)

giving

1+(1+\lambda_v(t_f))u(t_f)-g\lambda_v(t_f)=0,

(14)

since $v(t_f)=0$ .

Transversality¶

Both initial and terminal states are completely specified, so the transversality conditions provide no additional costate information.

Solution Strategy¶

The solution proceeds by:

Assume one of the admissible bang–bang structures.
Integrate dynamics on each segment.
Enforce continuity at the switching time.
Apply terminal conditions.
Verify the Hamiltonian condition.
Compare admissible solutions.

The state trajectories are piecewise:

velocity: linear,
altitude: quadratic.

Historical Significance¶

George Meditch’s 1964 paper showed how optimal control could be used to derive fuel-optimal lunar landing trajectories. This problem became one of the classical examples in aerospace optimal control.

Connection. The aircraft climb problem shows a different engineering strategy: reduce the model through a physically meaningful energy coordinate before optimizing.