Natural boundary condition

Remarks

Let $L$ be of class $C^2$ and let $y \in C^2([a, b])$ minimise $J$ on this class. For $\eta \in C^1([a, b])$ with $\eta(a) = 0$, integration by parts in the first variation gives $$0 = \delta J(y; \eta) = \int_a^b \Big( L_z - \frac{d}{dx} L_p \Big) \eta \, dx + L_p\big(b, y(b), y'(b)\big) \, \eta(b).$$ Variations with $\eta(b) = 0$ give the Euler–Lagrange equation, so the integral vanishes; a variation with $\eta(b) = 1$ then gives the natural boundary condition.

Example: let $J(y) = \int_0^1 \big( \frac{1}{2} y'^2 - y \big) \, dx$ with $y(0) = 0$ and $y(1)$ free. The Euler–Lagrange equation is $y'' = -1$ and the natural boundary condition is $y'(1) = 0$. The function $y(x) = x - x^2/2$ satisfies both: $y'' = -1$, $y(0) = 0$ and $y'(1) = 1 - 1 = 0$. Non-example: $y(x) = x$ has $y(0) = 0$, but $L_p = y' = 1$ at $x = 1$, so it fails the natural boundary condition.

In several variables, for $J(u) = \int_\Omega \big( \frac{1}{2} \abs{\nabla u}^2 - f u \big) \, dx$ with no condition on $\partial \Omega$, the natural boundary condition is $\partial u / \partial \nu = 0$, where $\nu$ is the outer normal.