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  "body": "\n## Problem\n\nLet $U \\subseteq \\R^m$ be a nonempty set, and let $t_0 \u003c t_1$ be fixed times.\nLet $f \\colon [t_0, t_1] \\times \\R^n \\times U \\to \\R^n$ be continuous, and let $S \\subseteq \\R^n \\times \\R^n$ be a closed set of allowed endpoint pairs.\nAn [[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 $(x(t_0), x(t_1)) \\in S$.\n\n1. 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 $u$.\n2. 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 $u$.\n\n## Remarks\n\nThe Mayer problem has only an endpoint cost, and the Lagrange problem has only a running cost.\nBoth are special cases of the Bolza problem, whose cost is the sum of the two.\nThey are equivalent to each other.\nA Lagrange problem becomes a Mayer problem with the extra state $y' = L(t, x, u)$, $y(t_0) = 0$, and the endpoint cost $y(t_1)$.\nA 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 $\\varphi(x(t_1)) = \\varphi(a) + \\int_{t_0}^{t_1} \\nabla\\varphi(x(t)) \\cdot x'(t) \\, dt$.\nThe Mayer form is convenient for existence proofs and for geometric arguments with reachable sets.\nThe Lagrange form is the classical form of the calculus of variations.\n\n## Explanations\n\n### L1 human\n\nIn a Mayer problem one only cares about where the system ends up, for example the final altitude of a rocket.\nIn a Lagrange problem one only cares about the total cost along the way, for example the fuel spent.\nA simple trick of bookkeeping, adding a counter for the cost so far, turns one kind of problem into the other.\n",
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