The [calculus of variations](/page/Calculus%20of%20Variations) studies the optimization of functionals—quantities that depend on entire functions or paths rather than just finite sets of variables. Where ordinary calculus asks "what value of $x$ minimizes $f(x)$?", the calculus of variations asks "what function $y(x)$ minimizes a given functional like $\int_a^b L(x, y, y') \, dx$?" This fundamental shift in perspective leads to some of the most powerful and elegant mathematics in analysis, with applications spanning classical mechanics, optics, differential geometry, and physics. This course develops the classical theory systematically, building from concrete variational problems to the abstract structures that unify and explain them.

The course begins by introducing functionals and the variational problem, then derives the Euler–Lagrange equation—the central [necessary condition for an extremum](/theorems/3520) of a functional. Subsequent chapters refine this picture: we handle boundary conditions and constraints, develop tests for distinguishing minima, maxima, and saddle points (via Legendre's condition and the second variation), and analyze the role of conjugate points in determining sufficiency. Along the way, we encounter the Weierstrass excess function, which characterizes *strong* extrema that remain optimal under perturbations not just in the function but in its derivative as well. These chapters form a complete sufficiency theory for classical extremal problems.

The course culminates in perspectives that connect the calculus of variations to deeper structures in mathematics and physics. Noether's theorem reveals that every symmetry of a variational problem yields a conservation law, unifying mechanics and geometry. The Hamiltonian formulation recasts the theory in terms of phase space rather than configuration space, and Hamilton–Jacobi theory provides a powerful method for solving variational problems and understanding the structure of solutions globally. The final chapter shows how Hamilton–Jacobi analysis completes the sufficiency story, tying together the classical necessary conditions of the middle chapters with a comprehensive existence and optimality framework.

# Introduction

What kind of mathematical object has a derivative when its input is an entire curve rather than a point in Euclidean space? This course studies that question in the classical setting: smooth curves, integral functionals, and the differential equations forced by extremality. The central theme is that a problem about minimising a number attached to a path often becomes a boundary value problem for an ordinary differential equation, and sometimes a first-order partial differential equation gives a global certificate of minimality.

The course is classical in two senses. First, the main unknowns are sufficiently smooth curves, so variations may be computed by differentiating under the integral sign and integrating by parts. Second, the theory is organised around the Euler--Lagrange equation, the second variation, Jacobi fields, conjugate points, Noether's theorem, and the Hamilton--Jacobi equation. Later courses in the series replace parts of this smooth picture by weak compactness, lower semicontinuity, relaxation, and Gamma-convergence.

## What the Course Is About

Which features of a minimisation problem survive when the variable is a function? In finite dimensions, a differentiable function $F: U \subseteq \mathbb R^n \to \mathbb R$ has a local extremum at an interior point $x_0$ only if $\nabla F(x_0)=0$. The calculus of variations asks for the analogue when the variable is a curve $y: [a,b] \to \mathbb R^n$ and the quantity to be extremised is usually an integral depending on $y$ and its derivative.

[definition: Integral Functional] Let $a<b$, let $U\subseteq \mathbb R^n$ be open, and let $L:[a,b]\times U\times \mathbb R^n\to \mathbb R$ be a smooth function. An integral functional associated to $L$ is a map \begin{align*} J[y] = \int_a^b L(x,y(x),y'(x))\,dx \end{align*} defined on a specified class of curves $y\in C^1([a,b];U)$ satisfying prescribed admissibility conditions. [/definition]

The function $L$ is called the Lagrangian. The variable $x$ often represents time, arc length, or a spatial parameter; the curve $y$ is the unknown; and $y'$ records the velocity or slope. The admissibility conditions are not decoration: fixed endpoints, free endpoints, constraints, and transversality conditions change the differential equations and boundary conditions that arise.

[example: Length Of A Plane Curve] For curves $y\in C^1([a,b];\mathbb R)$ with fixed endpoints, the Euclidean length of the graph $x\mapsto (x,y(x))$ is \begin{align*} J[y]=\int_a^b \sqrt{1+(y'(x))^2}\,dx. \end{align*} Here the Lagrangian is $L(x,y,p)=\sqrt{1+p^2}$, so its derivatives in the $y$ and $p$ variables are \begin{align*} \partial_y L(x,y,p)=0 \end{align*} and \begin{align*} \partial_p L(x,y,p)=\frac{p}{\sqrt{1+p^2}}. \end{align*} Thus the Euler--Lagrange equation for a smooth stationary curve becomes \begin{align*} \frac{d}{dx}\left(\frac{y'(x)}{\sqrt{1+(y'(x))^2}}\right)=0. \end{align*} This says that the function \begin{align*} \frac{y'(x)}{\sqrt{1+(y'(x))^2}} \end{align*} is constant on $[a,b]$. If $f(p)=p/\sqrt{1+p^2}$, then \begin{align*} f'(p)=\frac{\sqrt{1+p^2}-p\cdot p(1+p^2)^{-1/2}}{1+p^2} \end{align*} and the numerator equals \begin{align*} \sqrt{1+p^2}-\frac{p^2}{\sqrt{1+p^2}}=\frac{1+p^2-p^2}{\sqrt{1+p^2}}=\frac{1}{\sqrt{1+p^2}}. \end{align*} Therefore \begin{align*} f'(p)=\frac{1}{(1+p^2)^{3/2}}>0. \end{align*} So $f$ is strictly increasing, and $f(y'(x))$ being constant forces $y'(x)$ itself to be constant. Hence $y(x)=mx+c$ for constants $m,c$, and the extremals for the length functional are straight line segments. [/example]

This first example explains the basic shape of the subject: a geometric or physical quantity is encoded by $L$, a class of admissible curves is fixed, and stationarity is translated into equations. The next issue is that a stationary curve need not be a minimiser, so the course distinguishes necessary conditions from sufficient conditions throughout.

## From Variations To Differential Equations

How does a small perturbation of a curve produce an equation? If $y$ is admissible and $h$ is an allowed variation direction, the perturbed curves $y+\varepsilon h$ generate a one-variable function $\varepsilon\mapsto J[y+\varepsilon h]$. The derivative at $\varepsilon=0$ is the first variation, and an interior extremum must make this derivative vanish for every allowed $h$.

[definition: First Variation] Let $\mathcal A\subset C^1([a,b];\mathbb R^n)$ be an admissible class of curves, and let $J:\mathcal A\to\mathbb R$ be a functional. If $y\in\mathcal A$ and $h$ is an admissible variation direction at $y$, the first variation of $J$ at $y$ in the direction $h$ is \begin{align*} \delta J[y;h] = \frac{d}{d\varepsilon}\Big|_{\varepsilon=0} J[y+\varepsilon h], \end{align*} whenever this derivative exists. [/definition]

This definition is the infinite-dimensional analogue of a directional derivative. In the basic fixed-endpoint problem, $h(a)=h(b)=0$, so [integration by parts](/theorems/210) transfers derivatives from $h$ onto the coefficients involving $L$. The vanishing of $\delta J[y;h]$ for all such $h$ then forces the Euler--Lagrange equation.

[quotetheorem:6986]

[citeproof:6986]

The theorem gives only stationarity, not minimality. The hypotheses about admissible directions matter because a perturbation must preserve the endpoint and constraint conditions; otherwise the one-variable test is being applied along curves outside the problem. Much of the first half of the course develops the consequences of this stationarity condition, while the second half asks when stationarity plus additional hypotheses proves that a curve is actually minimising; the later example of a stationary curve with negative second variation shows concretely why the first variation cannot settle minimality by itself.

[example: Brachistochrone Setup] Choose coordinates with $x$ horizontal and $y$ measuring downward vertical distance from the release point, so the bead starts at $y=0$ and admissible curves satisfy $y(x)>0$ for $x>0$. Conservation of mechanical energy gives the speed at height $y$ as $v=\sqrt{2gy}$, and the length element along the graph is $\sqrt{1+(y'(x))^2}\,dx$, so the travel-time functional is \begin{align*} T[y]=\int_0^b \frac{\sqrt{1+(y'(x))^2}}{\sqrt{2g\,y(x)}}\,dx. \end{align*} Thus \begin{align*} L(y,p)=\frac{\sqrt{1+p^2}}{\sqrt{2gy}}. \end{align*} Since $L$ has no explicit $x$-dependence, the Euler--Lagrange equation implies that $L-p\partial_pL$ is constant along any smooth extremal. Indeed, \begin{align*} \frac{d}{dx}\bigl(L-y'\partial_pL\bigr)=\partial_yL\,y'+\partial_pL\,y''-y''\partial_pL-y'\frac{d}{dx}\partial_pL. \end{align*} The middle terms cancel, and the Euler--Lagrange equation gives $\frac{d}{dx}\partial_pL=\partial_yL$, so \begin{align*} \frac{d}{dx}\bigl(L-y'\partial_pL\bigr)=\partial_yL\,y'-y'\partial_yL=0. \end{align*} For the brachistochrone Lagrangian, \begin{align*} \partial_pL(y,p)=\frac{p}{\sqrt{2gy}\sqrt{1+p^2}}. \end{align*} Therefore \begin{align*} L-p\partial_pL=\frac{\sqrt{1+p^2}}{\sqrt{2gy}}-\frac{p^2}{\sqrt{2gy}\sqrt{1+p^2}}. \end{align*} Putting the terms over the common denominator $\sqrt{2gy}\sqrt{1+p^2}$ gives \begin{align*} L-p\partial_pL=\frac{1+p^2-p^2}{\sqrt{2gy}\sqrt{1+p^2}}. \end{align*} Hence \begin{align*} L-p\partial_pL=\frac{1}{\sqrt{2gy}\sqrt{1+p^2}}. \end{align*} Along an extremal this equals a positive constant $C$, so with $p=y'$ we get \begin{align*} 2g\,y(x)\bigl(1+(y'(x))^2\bigr)=\frac{1}{C^2}. \end{align*} Writing $a=1/(2gC^2)>0$, this becomes \begin{align*} y(x)\bigl(1+(y'(x))^2\bigr)=a. \end{align*} Solving for $y'$ on a descending branch gives \begin{align*} y'(x)=\sqrt{\frac{a-y(x)}{y(x)}}. \end{align*} Equivalently, \begin{align*} \frac{dx}{dy}=\sqrt{\frac{y}{a-y}}. \end{align*} Set \begin{align*} y=\frac{a}{2}(1-\cos\theta). \end{align*} Then \begin{align*} a-y=\frac{a}{2}(1+\cos\theta). \end{align*} Also \begin{align*} \frac{dy}{d\theta}=\frac{a}{2}\sin\theta. \end{align*} Using $1-\cos\theta=2\sin^2(\theta/2)$ and $1+\cos\theta=2\cos^2(\theta/2)$, \begin{align*} \sqrt{\frac{y}{a-y}}=\tan(\theta/2). \end{align*} Thus \begin{align*} \frac{dx}{d\theta}=\frac{dx}{dy}\frac{dy}{d\theta}=\tan(\theta/2)\cdot \frac{a}{2}\sin\theta. \end{align*} Since $\tan(\theta/2)=\sin\theta/(1+\cos\theta)$, this is \begin{align*} \frac{dx}{d\theta}=\frac{a}{2}\frac{\sin^2\theta}{1+\cos\theta}. \end{align*} Using $\sin^2\theta=(1-\cos\theta)(1+\cos\theta)$, \begin{align*} \frac{dx}{d\theta}=\frac{a}{2}(1-\cos\theta). \end{align*} Integrating from the initial point gives \begin{align*} x=\frac{a}{2}(\theta-\sin\theta). \end{align*} Together with \begin{align*} y=\frac{a}{2}(1-\cos\theta), \end{align*} this is a cycloid. Thus the Euler--Lagrange equation does not produce a straight chord for the fastest descent problem; it selects cycloidal arcs, with the parameter $a$ and endpoint value of $\theta$ determined by the prescribed final point. [/example]