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  "body": "\n## Statement\n\nLet $a \u003c b$ and let $L \\colon [a, b] \\times \\R^n \\times \\R^n \\to \\R$, written $L(x, y, p)$, be of class $C^1$.\nLet $J(y) = \\int_a^b L(x, y, y') \\, dx$ on piecewise $C^1$ functions $y \\colon [a, b] \\to \\R^n$ with fixed values at $a$ and $b$.\nLet $y_0$ be a piecewise $C^1$ [[strong local minimiser|WMPCR723MJ]] of $J$ in this class.\nLet $c \\in (a, b)$ be a corner of $y_0$, with one-sided derivatives $p_- = y_0'(c^-)$ and $p_+ = y_0'(c^+)$.\nThen\n$$L_p(c, y_0(c), p_-) = L_p(c, y_0(c), p_+),$$\n$$L(c, y_0(c), p_-) - p_- \\cdot L_p(c, y_0(c), p_-) = L(c, y_0(c), p_+) - p_+ \\cdot L_p(c, y_0(c), p_+).$$\n\n## Remarks\n\nThe first condition comes from variations $y_0 + \\varepsilon h$ with $h$ supported near $c$.\nThe du Bois-Reymond form of the [[Euler–Lagrange equation|PS0VV149ET]] says that $L_p(x, y_0, y_0')$ equals a continuous function of $x$, so it has no jump at $c$.\nThe second condition comes from moving the corner point sideways, which is a variation of the independent variable $x$.\n\nExample: let $J(y) = \\int_0^2 (y'^2 - 1)^2 \\, dx$ with $y(0) = y(2) = 0$.\nThe zigzag $y_0(x) = \\min(x, 2 - x)$ has $J(y_0) = 0$, so it is a minimiser, with a corner at $1$ where $p_- = 1$ and $p_+ = -1$.\nHere $L_p = 4p(p^2 - 1)$ and $L - p L_p = (p^2 - 1)^2 - 4p^2(p^2 - 1)$, and both vanish at $p = \\pm 1$, as the theorem requires.\n\nIf $L_{pp} \u003e 0$ everywhere, the map $p \\mapsto L_p(x, y, p)$ is injective when $n = 1$, and the first condition forces $p_- = p_+$.\nSo for such Lagrangians minimisers have no corners.\n\n## Explanations\n\n### L1 human\n\nSometimes the best curve has a sharp bend, a corner.\nThe corner conditions say which bends are allowed.\nTwo quantities built from the integrand, often called the momentum and the energy, must take the same value just before and just after the corner.\n\n### L2 human\n\nThe Euler–Lagrange equation holds on each smooth piece of a broken extremal, but it says nothing about how the pieces join.\nThe Weierstrass–Erdmann conditions supply the missing junction conditions, two of them, which match the freedom in the location and the slopes at a corner.\nIn Hamiltonian terms they say that the momentum $L_p$ and the Hamiltonian $p \\cdot L_p - L$ are continuous along the extremal.\nThey explain why problems with Lagrangians that are not convex in $p$ often have broken minimisers.\n",
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