Euler–Lagrange equation

Remarks

If $\Omega$ is bounded and $u \in C^2(\overline{\Omega})$ minimises $J$ among $C^2(\overline{\Omega})$ functions with the same boundary values, then $u$ solves the Euler–Lagrange equation. Indeed $\frac{d}{dt} J(u + t\varphi)\big|_{t=0} = 0$ for every $\varphi \in C^\infty_c(\Omega)$; integrating by parts gives $\int_\Omega (-\operatorname{div} L_p + L_z) \varphi \, dx = 0$ for all such $\varphi$, so the integrand vanishes. The weak form $\int_\Omega \big( L_p(x, u, \nabla u) \cdot \nabla \varphi + L_z(x, u, \nabla u) \varphi \big) \, dx = 0$ for all $\varphi \in C^\infty_c(\Omega)$ makes sense for less regular $u$.

Example: for $L(x, z, p) = \frac{1}{2}\abs{p}^2 - f(x) z$ one has $L_p = p$ and $L_z = -f$, so the Euler–Lagrange equation is Poisson's equation $-\Delta u = f$.

Example: for $n = 1$ and $L(x, z, p) = \sqrt{1 + p^2}$, one gets $\frac{d}{dx}\big(u' / \sqrt{1 + u'^2}\big) = 0$, so $u'$ is constant and the graph is a straight line.

Non-example: a solution of the Euler–Lagrange equation need not be a minimiser. For $n = 1$, $\Omega = (0, 2\pi)$ and $L = \frac{1}{2}(p^2 - z^2)$, the equation is $u'' + u = 0$, solved by $u = 0$; but $J(t \sin(x/2)) = \frac{t^2}{2} \int_0^{2\pi} \big(\frac{1}{4}\cos^2(x/2) - \sin^2(x/2)\big) \, dx = -\frac{3\pi t^2}{8} < 0 = J(0)$ for $t \ne 0$, and $t\sin(x/2)$ has the same boundary values $0$.