---
tag: 1J8TYT7C49
kind: problem
name: Mayer and Lagrange problems
msc: [49K15, 49J15]
uses: [T3V6ZNAMDN, ZPDTVZQN5K, 8M31QKB0GW, 1EQM0DJ4NC]
about: [T3V6ZNAMDN]
sources:
  - url: https://en.wikipedia.org/wiki/Optimal_control
history:
  posed_by: "Joseph-Louis Lagrange and Adolph Mayer"
---

## Problem

Let $U \subseteq \R^m$ be a nonempty set, and let $t_0 < t_1$ be fixed times.
Let $f \colon [t_0, t_1] \times \R^n \times U \to \R^n$ be continuous, and let $S \subseteq \R^n \times \R^n$ be a closed set of allowed endpoint pairs.
An [[admissible control|8M31QKB0GW]] is a measurable map $u \colon [t_0, t_1] \to U$ whose trajectory $x$ of the [[control system|ZPDTVZQN5K]] $x'(t) = f(t, x(t), u(t))$ is absolutely continuous on $[t_0, t_1]$ and satisfies $(x(t_0), x(t_1)) \in S$.

1. The *Mayer problem*: given a continuous function $\varphi \colon \R^n \times \R^n \to \R$, minimize $J(u) = \varphi(x(t_0), x(t_1))$ over all admissible controls $u$.
2. The *Lagrange problem*: given a continuous function $L \colon [t_0, t_1] \times \R^n \times U \to \R$, minimize the [[functional|1EQM0DJ4NC]] $J(u) = \int_{t_0}^{t_1} L(t, x(t), u(t)) \, dt$ over all admissible controls $u$.

## Remarks

The Mayer problem has only an endpoint cost, and the Lagrange problem has only a running cost.
Both are special cases of the Bolza problem, whose cost is the sum of the two.
They are equivalent to each other.
A Lagrange problem becomes a Mayer problem with the extra state $y' = L(t, x, u)$, $y(t_0) = 0$, and the endpoint cost $y(t_1)$.
A Mayer problem with $\varphi$ of class $C^1$ that depends only on $x(t_1)$, and a fixed initial state $x(t_0) = a$, becomes a Lagrange problem with $L(t, x, u) = \nabla \varphi(x) \cdot f(t, x, u)$, since $\varphi(x(t_1)) = \varphi(a) + \int_{t_0}^{t_1} \nabla\varphi(x(t)) \cdot x'(t) \, dt$.
The Mayer form is convenient for existence proofs and for geometric arguments with reachable sets.
The Lagrange form is the classical form of the calculus of variations.

## Explanations

### L1 human

In a Mayer problem one only cares about where the system ends up, for example the final altitude of a rocket.
In a Lagrange problem one only cares about the total cost along the way, for example the fuel spent.
A simple trick of bookkeeping, adding a counter for the cost so far, turns one kind of problem into the other.