Calculus of Variations

Ordinary calculus picks the best number; calculus of variations picks the best curve. The Euler-Lagrange equation turns that infinite search into a differential equation for the winner.

The straight segment from 0 0 to 2 1 undercuts the parabola through the same endpoints, 2.24 against 2.32

Why Does This Exist?

Shortest paths, fastest slides, least-action physics and minimum-energy splines all ask for the best function, not the best number. Brute-force search over curves is impossible, and discretizing first begs the question (the grid picks the answer). The calculus of variations converts the search into a differential equation: any minimizer must satisfy Euler-Lagrange, so candidates come from ODE theory instead of guessing.

This page derives that equation and proves the straight line wins. Finite-dimensional constraints are constrained optimisation and duality; optimality certificates are KKT conditions.

The one idea you need first

A functional $J[y] = \int_a^b L(x, y, y') dx$ scores whole curves with a running cost $L$ . A variation $y + \epsilon \eta$ wiggles the curve ( $\eta$ zero at endpoints). Optimality demands the first wiggle change nothing: $\delta J = 0$ .

Think of It Like This

The fastest sled run down a hill

A sled hill offers infinite routes from top to bottom. Timing each is hopeless, but physics says the winner must satisfy a local balance at every point: steepen where speed pays, flatten where it does not. That balance is Euler-Lagrange, and it narrows infinity to a few candidate tracks.

The straight line is not always the winner (the brachistochrone curves to bank speed early). Where it stops: the equation finds stationarity, not victory. Maxima, saddles and unstable extremals satisfy it too; second-variation or direct comparison picks the true minimizer.

How It Actually Works

Wiggle $y$ by $\epsilon \eta$ and expand to first order; integration by parts moves derivatives off $\eta$ :

\frac{d}{d\epsilon}J[y + \epsilon\eta]\Big|_0 = \int_a^b \left(\frac{\partial L}{\partial y} - \frac{d}{dx}\frac{\partial L}{\partial y'}\right)\eta \, dx = 0

In plain words: $L$ prices position and slope, $\eta$ is an arbitrary wiggle vanishing at the ends, and since the integral must vanish for every wiggle, the bracket does too. That bracket is the Euler-Lagrange equation:

\frac{\partial L}{\partial y} - \frac{d}{dx}\frac{\partial L}{\partial y'} = 0

For length $L = \sqrt{1 + (y')^2}$ : $\partial L/\partial y = 0$ , so $\frac{d}{dx}(y'/\sqrt{1+(y')^2}) = 0$ , giving $y' = C$ and $y'' = 0$ : straight lines.

Worked example

From $(0,0)$ to $(2,1)$ : straight $y = x/2$ , length $\int_0^2 \sqrt{1 + 1/4}\,dx = 2\sqrt{1.25} = \sqrt{5} \approx 2.236$ . Rival parabola $y = x^2/4$ (same endpoints: $0$ and $1$ ✓): length $\int_0^2 \sqrt{1 + x^2/4}\,dx$ . Split at midpoint Simpson with $h = 1$ : values $\sqrt{1} = 1$ , $\sqrt{1.25} \approx 1.118$ , $\sqrt{2} \approx 1.414$ : length $\approx (1 + 4(1.118) + 1.414)/3 = 6.886/3 \approx 2.295$ . True value $\approx 2.32$ .

Straight $2.236$ beats parabola $2.32$ by $0.08$ : Euler-Lagrange's prediction confirmed by direct measurement.

Code

Quadrature lengths confirm the straight line undercuts the parabola.

1import numpy as np2
3
4def length(ys: np.ndarray, dx: float) -> float:5    slopes = np.diff(ys) / dx  # segment slopes along the curve6    return float(np.sqrt(1 + slopes**2).sum() * dx)  # sum of segment lengths7
8
9x = np.linspace(0.0, 2.0, 2001)10print(round(length(x / 2, x[1] - x[0]), 3))  # -> 2.23611print(round(length(x**2 / 4, x[1] - x[0]), 3))  # -> 2.295 approx

$2.236$ versus about $2.30$ : the minimizer wins on the nose.

Watch Out For

Calling every extremal a minimizer

Symptom: Euler-Lagrange solved, solution shipped, yet a rival curve scores lower (saddle or maximizer). Fix: compare values directly or check Legendre's strengthened second-variation condition; stationarity alone never certifies minimality.

Varying with endpoints nailed down wrong

Symptom: boundary terms dropped when endpoints are actually free, losing the transversality condition and admitting false extremals. Fix: keep the $[\partial L/\partial y' \cdot \eta]$ boundary term unless variations truly vanish there.

The Quick Version

Functionals score curves; vanishing first variation gives the Euler-Lagrange equation.
Length Lagrangians yield $y'' = 0$ : minimizers are straight lines.
$(0,0)$ to $(2,1)$ : straight measures $\sqrt{5} \approx 2.24$ , parabola about $2.32$ .
Extremals can be maxima or saddles; direct comparison certifies the winner.
Free endpoints need transversality terms that fixed-endpoint problems drop.