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  "body": "\n## Statement\n\nLet $L \\colon \\R \\times \\R \\times \\R \\to \\R$, written $L(x, z, p)$, be of class $C^2$, with [[partial derivative|09EGC6F8PT]] $L_p$ in $p$.\nLet $\\psi \\colon \\R \\to \\R$ be of class $C^1$, and fix $a, A \\in \\R$.\nConsider pairs $(y, b)$ with $b \u003e a$, $y \\in C^2([a, b])$, $y(a) = A$ and $y(b) = \\psi(b)$, and set $J(y, b) = \\int_a^b L(x, y, y') \\, dx$.\nSuppose that $(y_0, b_0)$ minimises $J$ among all such pairs.\nThen:\n\n1. $y_0$ solves the [[Euler–Lagrange equation|PS0VV149ET]] $\\frac{d}{dx} L_p(x, y_0, y_0') = L_z(x, y_0, y_0')$ on $[a, b_0]$;\n2. at the free end, the *transversality condition*\n$$\\Big[ L(x, y_0, y_0') + \\big(\\psi'(x) - y_0'(x)\\big) L_p(x, y_0, y_0') \\Big]_{x = b_0} = 0$$\nholds.\n\n## Remarks\n\nThe proof computes the [[first variation|5B8NFYB2XQ]] of $J$ along a family $(y_t, b_t)$ of competitors with $y_t(b_t) = \\psi(b_t)$.\nVariations that fix both ends give the Euler–Lagrange equation.\nMoving the end along the curve $y = \\psi(x)$ gives a boundary term, $L \\, \\delta b + L_p (\\delta y - y' \\delta b)$ at $b_0$, with $\\delta y = \\psi' \\delta b$; it must vanish, which is the transversality condition.\n\nSpecial cases.\nIf instead $b$ is fixed and $y(b)$ is free, the end moves on the vertical line $x = b$, and the same computation gives the [[natural boundary condition|QTRPNFDFJ8]] $L_p = 0$ at $b$.\nIf $L(x, z, p) = f(x, z) \\sqrt{1 + p^2}$ with $f(b_0, y_0(b_0)) \\ne 0$, the condition simplifies to $\\psi'(b_0) \\, y_0'(b_0) = -1$: the extremal meets the curve $y = \\psi(x)$ at a right angle.\n\nExample: for $f = 1$, $J$ is arc length, and the condition says that the shortest segment from the point $(a, A)$ to a curve meets the curve orthogonally.\n\n## Explanations\n\n### L1 human\n\nWhen the end of a curve may slide along a given line or curve, the best curve must meet that curve in a special way.\nFor the shortest path from a point to a curve, the special way is a right angle.\nFor other functionals, the angle is fixed by a formula involving the integrand.\n\n### L2 human\n\nThe transversality condition replaces the missing boundary value.\nThe Euler–Lagrange equation has order two and one end is fixed, so one more condition fixes the extremal; then $y(b_0) = \\psi(b_0)$ fixes the unknown $b_0$.\nIn optimal control, the [[Pontryagin maximum principle|7RPV2T29CF]] has transversality conditions of the same kind on the costate at a free final state or final time.\n",
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