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  "body": "\n## Definition\n\nLet $a \u003c b$, let $G \\subseteq \\R \\times \\R^n$ be open, and let $L \\colon G \\times \\R^n \\to \\R$ be continuous, written $L(t, x, v)$.\nFor a $C^1$ curve $y \\colon [a, b] \\to \\R^n$ with graph in $G$, let $J(y) = \\int_a^b L(t, y(t), y'(t)) \\, dt$, a functional of the [[calculus of variations|PQ25NHV0KS]].\nLet $\\bar x$ be a $C^1$ curve with graph in $G$.\n*Carathéodory's royal road* is the following method to show that $\\bar x$ minimises $J$ among such curves with the same endpoints.\nOne looks for a $C^1$ function $S \\colon G \\to \\R$ and a continuous map $\\psi \\colon G \\to \\R^n$ such that:\n\n1. $L(t, x, v) - S_t(t, x) - \\nabla_x S(t, x) \\cdot v \\ge 0$ for all $(t, x) \\in G$ and $v \\in \\R^n$;\n2. $L(t, x, \\psi(t, x)) - S_t(t, x) - \\nabla_x S(t, x) \\cdot \\psi(t, x) = 0$ for all $(t, x) \\in G$;\n3. $\\bar x'(t) = \\psi(t, \\bar x(t))$ for all $t \\in [a, b]$.\n\nThe function $L^*(t, x, v) = L(t, x, v) - S_t(t, x) - \\nabla_x S(t, x) \\cdot v$ is the *equivalent Lagrangian* of $L$ with respect to $S$.\n\n## Remarks\n\nThe method works because of this identity, valid for every admissible curve $y$:\n$$J(y) = \\int_a^b L^*(t, y(t), y'(t)) \\, dt + S(b, y(b)) - S(a, y(a)).$$\nThe boundary terms depend only on the endpoints.\nBy clauses 1 to 3, the integral of $L^*$ is $\\ge 0$ for every $y$ and $= 0$ for $\\bar x$, so $J(y) \\ge J(\\bar x)$.\n\nClauses 1 and 2 say that $\\min_v L^*(t, x, v) = 0$, with the minimum at $v = \\psi(t, x)$.\nWith $H(t, x, p) = \\sup_v (p \\cdot v - L(t, x, v))$, this means $S_t + H(t, x, \\nabla_x S) = 0$, the [[Hamilton–Jacobi equation|5TRXQKP7DY]].\n\nExample: $n = 1$, $G = \\R^2$, $L = \\frac{1}{2} v^2$, and $\\bar x(t) = t$ on $[0, 1]$.\nTake $S(t, x) = x - t/2$ and $\\psi = 1$.\nThen $L^* = \\frac{1}{2} v^2 + \\frac{1}{2} - v = \\frac{1}{2} (v - 1)^2$, which is $\\ge 0$ and vanishes at $v = 1$, and $\\bar x' = 1$.\nSo $\\bar x(t) = t$ minimises $\\int_0^1 \\frac{1}{2} y'^2 \\, dt$ among $C^1$ curves from $(0, 0)$ to $(1, 1)$.\n\nNon-example: in the same problem, $S = 0$ with $\\psi = 1$ fails clause 2, since $L(t, x, 1) = \\frac{1}{2} \\ne 0$.\n\n## Explanations\n\n### L1 human\n\nThe method changes the integrand by a total derivative, which changes the integral only by endpoint terms.\nIf the new integrand is never negative and is zero along the candidate curve, the candidate wins.\n",
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