Calculus of Variations

Stationarity, Euler-Lagrange residuals, and the brachistochrone as things you can touch.

Calculus of variations optimizes over functions instead of numbers. A curve $y(x)$ is plugged into a functional

$$ J[y] = \int_a^b L(x,y,y')\,dx, $$

and the question is whether small perturbations of the whole curve make $J$ go up, down, or stay flat to first order. The widgets below put three views next to each other: the global perturbation test $\delta J=0$, the local Euler-Lagrange residual, and a physical race where the optimum is a cycloid.

One sentence: a stationary curve is not necessarily the lowest, shortest, or steepest descent; it is one whose first-order change in the functional vanishes for every allowed perturbation.

1. First variation microscope

The base curve is a sine-series path with fixed endpoints. The perturbation $\eta(x)$ also vanishes at the endpoints, so $y_\epsilon(x)=y(x)+\epsilon\eta(x)$ respects the boundary conditions. The functional is a loaded string energy,

$$ J[y] = \int_0^1 \left(\frac12 y'(x)^2 - 4y(x)\right)dx. $$

The stationary curve satisfies $y''=-4$, hence $y^\star(x)=2x(1-x)$. The right panel plots $J[y+\epsilon\eta]$ against $\epsilon$. If the tangent at zero is not flat, the current curve still has a first-order way to improve.

What this functional is. $J[y] = \int_0^1 \bigl(\tfrac12 y'^2 - 4y\bigr)dx$ is the small-deflection elastic energy of a string pinned at both ends under a uniform downward load of intensity $4$. The $\tfrac12 y'^2$ term is elastic stretch energy, and $-4y$ is the work done by gravity as the string sags into shape $y(x)$. The Euler-Lagrange equation $y''=-4$ is exactly the static balance: curvature opposing load. So a stationary curve here is a string in equilibrium, and "dropping" toward the optimum (Figure 1's Relax button) is what the physical string would do under damping.

A second perspective: stationary means flat in every direction at once. The readout lists $dJ/d\epsilon|_0$ for all four perturbation modes simultaneously. A non-stationary curve will show at least one mode with a non-zero slope; the stationary curve flattens all of them together.

Drag a point on the left panel to deform the curve at that $x$, or drag horizontally on the right panel to scrub $\epsilon$.

Figure 1 · First variation: J[y + epsilon eta]

curve coefficient c1 0.25

curve coefficient c2 0.42

curve coefficient c3 -0.25

perturbation epsilon 0.45

perturbation mode 2

2. Euler-Lagrange residual heatmap

For the same functional, the Euler-Lagrange residual is

$$ R(x) = \frac{\partial L}{\partial y} - \frac{d}{dx}\frac{\partial L}{\partial y'} = -4 - y''(x). $$

A stationary curve has $R(x)=0$ at every interior point. This is the local diagnostic version of the first-variation test above: red means that piece of the curve is still violating the Euler-Lagrange equation. Drag a point on the curve to deform it at that $x$ and watch the residual color shift in response.

Figure 2 · Local Euler-Lagrange residual along the curve

large residual near zero residual Euler-Lagrange solution

3. Brachistochrone race

The brachistochrone asks for the fastest frictionless track between two points under gravity. The shortest path is not fastest: an initially steep curve lets the bead build speed early. The time functional is

$$ T[y] = \int_{x_A}^{x_B} \sqrt{\frac{1+y'(x)^2}{2g\,y(x)}}\,dx, $$

where $y$ is the vertical drop below the start. Drag the two control points. The bead races your cubic Bezier track against the cycloid with the same endpoints, and the histogram records every curve you try.

Figure 3 · Race your curve against the cycloid

user curve cycloid optimum tried curve times

What these three views share

The first-variation plot asks whether $J[y+\epsilon\eta]$ has a flat tangent at $\epsilon=0$. The residual heatmap asks whether the Euler-Lagrange expression vanishes point by point. The brachistochrone shows the same variational logic in a physical system: among all admissible tracks, the cycloid is the curve where the time functional is stationary.

What next

Variational ideas reappear in probability as optimization over distributions.

Bayesian inference

Free Energy & Variational Inference

See the ELBO identity that turns posterior approximation into a variational optimization problem.

Mean-field VI

Exact Posterior vs Mean-Field VI

Watch factorized variational Bayes underestimate uncertainty in Bayesian linear regression.

Information

KL Divergence

Study the directed mismatch measure that drives variational Bayes.