Weierstrass–Erdmann corner conditions

Remarks

The first condition comes from variations $y_0 + \varepsilon h$ with $h$ supported near $c$. The du Bois-Reymond form of the Euler–Lagrange equation says that $L_p(x, y_0, y_0')$ equals a continuous function of $x$, so it has no jump at $c$. The second condition comes from moving the corner point sideways, which is a variation of the independent variable $x$.

Example: let $J(y) = \int_0^2 (y'^2 - 1)^2 \, dx$ with $y(0) = y(2) = 0$. The 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 $p_+ = -1$. Here $L_p = 4p(p^2 - 1)$ 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.

If $L_{pp} > 0$ everywhere, the map $p \mapsto L_p(x, y, p)$ is injective when $n = 1$, and the first condition forces $p_- = p_+$. So for such Lagrangians minimisers have no corners.

Going deeper

For a student

The Euler–Lagrange equation holds on each smooth piece of a broken extremal, but it says nothing about how the pieces join. The Weierstrass–Erdmann conditions supply the missing junction conditions, two of them, which match the freedom in the location and the slopes at a corner. In Hamiltonian terms they say that the momentum $L_p$ and the Hamiltonian $p \cdot L_p - L$ are continuous along the extremal. They explain why problems with Lagrangians that are not convex in $p$ often have broken minimisers.