Euler-Lagrange equation
/ OY-ler la-GRAHNZH /
If a curve makes a functional smallest or stationary, the Euler-Lagrange equation is the differential equation that curve must obey. It is the calculus-of-variations counterpart of the ordinary rule 'set the derivative to zero'. Where single-variable calculus answers 'which point?' with an algebraic equation, the calculus of variations answers 'which curve?' with this differential equation. Solving it (with the boundary conditions) hands you the candidate optimal path.
For a functional J[y] = integral from a to b of L(x, y, y') dx, the equation reads, in words: the x-derivative of (the partial of L with respect to y') equals the partial of L with respect to y. In compact symbols, d/dx of (dL/dy') = dL/dy, or equivalently dL/dy minus d/dx of (dL/dy') = 0. It comes straight out of the first variation: you wiggle the curve, integrate the y' term by parts to expose eta alone, and the fundamental lemma lets you drop the integral. Because the d/dx unpacks by the chain rule into terms with y'' , this is generally a second-order differential equation for y(x), to be solved with the two endpoint conditions.
This single equation is the workhorse behind nearly every classic variational result and behind all of analytical mechanics. Feed it the arc-length integrand and it returns straight lines; feed it the slide-time integrand and it returns the brachistochrone cycloid; feed it a mechanical Lagrangian (kinetic minus potential energy) and it returns Newton's laws of motion. An honest caveat: a solution of the Euler-Lagrange equation is only an extremal — a stationary path. It is a necessary condition for an optimum, not a guarantee of one; you still need the second variation, or physical or convexity arguments, to confirm a genuine minimum exists and is attained.
For arc length, L = sqrt(1 + y'^2) has dL/dy = 0, so the equation says d/dx of (dL/dy') = 0, i.e. y'/sqrt(1+y'^2) is constant. A constant slope means y' is constant, so y is a straight line — the shortest path between two points, derived rather than assumed.
Plug arc length into Euler-Lagrange and the straight line falls out as the shortest path.
When L has no explicit dependence on x, do not grind through the full second-order equation — the Beltrami identity gives a first integral that is far easier. And remember: a solution is an extremal, not automatically a minimum.