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  "body": "\n## Statement\n\nLet $a \u003c b$, and let $L \\colon \\R \\times \\R \\to \\R$, written $L(z, p)$, be a function of class $C^2$ that does not depend on the independent variable $x$.\nWrite $L_z$ and $L_p$ for its [[partial derivatives|09EGC6F8PT]].\nLet $y \\in C^2([a, b])$ solve the [[Euler–Lagrange equation|PS0VV149ET]]\n$$\\frac{d}{dx} L_p\\big(y(x), y'(x)\\big) = L_z\\big(y(x), y'(x)\\big) \\quad \\text{on } [a, b].$$\nThen there is a constant $C$ with\n$$L\\big(y(x), y'(x)\\big) - y'(x) \\, L_p\\big(y(x), y'(x)\\big) = C \\quad \\text{for all } x \\in [a, b].$$\n\n## Remarks\n\nThe proof is one differentiation.\nThe derivative of $L(y, y') - y' L_p(y, y')$ is $L_z y' + L_p y'' - y'' L_p - y' \\frac{d}{dx} L_p = y' \\big(L_z - \\frac{d}{dx} L_p\\big)$, which is $0$ by the Euler–Lagrange equation.\n\nExample: for the [[minimal surface of revolution|AEG6QWEFRG]] one has $L(z, p) = z \\sqrt{1 + p^2}$.\nThen $L - p L_p = z / \\sqrt{1 + p^2}$, so $y / \\sqrt{1 + y'^2} = C$.\nFor $C \u003e 0$ the solutions are the catenaries $y(x) = C \\cosh\\big((x - d)/C\\big)$.\n\nA partial converse holds: if the quantity is constant and $y'(x) \\ne 0$, then the Euler–Lagrange equation holds at $x$.\nSo constant solutions $y' = 0$ of the identity must be checked separately.\n\n## Explanations\n\n### L1 human\n\nWhen the integrand of a variational problem does not depend on position along the axis, one quantity stays constant along every solution.\nThis replaces a second-order equation by a first-order one, which is often easy to solve.\nIt is how the brachistochrone and the hanging chain are solved by hand.\n\n### L2 human\n\nIn mechanics, with $x$ as time and $L$ the Lagrangian, the quantity $y' L_p - L$ is the energy, the Hamiltonian.\nThe identity is then conservation of energy for a time-independent Lagrangian, a special case of [[Noether's theorem|XHPDM5T2JG]] for time translations.\nIt is used in the [[brachistochrone problem|ZWQ5JBR6XF]] and the [[catenary problem|8REBFQD6E0]].\n",
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