Part B · the action · spine

The brachistochrone, and how to differentiate with respect to a curve

A number that depends on a whole curve has a derivative too, and setting it to zero turns "nature extremises this integral" into a differential equation.

needs 04~2½ h slides + 6 h problemsproblems: ./mr problems 11

Everything so far has minimised over numbers. Unit 04's Lagrange multipliers found the lowest point of a potential on a surface: finitely many coordinates, finitely many equations. This unit minimises over curves. The unknown is a function, the space it lives in is infinite-dimensional, and the stationarity condition is no longer a finite list of equations but a differential equation.

The physics that forces it is old and was never optional. Fermat (1662) said light takes the quickest route; Johann Bernoulli (1696) asked which wire a bead descends fastest; Galileo, Huygens and Jacob Bernoulli all asked what shape a hanging chain takes. Each is an integral over a curve, to be made as small as possible. The common machine that answers all three is what gets built here.

The pay-off is Part B's whole thesis. Unit 12 shows that Newton's equations are the Euler–Lagrange equations of $\int (T - V)$, which makes them invariant under any change of coordinates at all; unit 13 turns symmetries of the same integral into conservation laws in three lines; unit 29 runs the identical argument over a region of spacetime; unit 32 makes free fall the extremal of arc length; and unit 35 derives the Einstein field equations by varying one integral over a four-dimensional manifold. All of that is this unit's derivation, repeated in richer settings.

How to read this deck

One problem runs through it, with numbers: a bead sliding from $(0,0)$ to $(2\ \mathrm m, 1\ \mathrm m)$ under gravity. Section 1 writes down the integral, section 2 builds the equation, section 3 solves it, section 4 handles the constraints and free ends the physics needs, section 5 asks whether the answer is really a minimum, and section 6 asks whether a minimum exists at all.

Proofs are on the slides where they are one line, and in these notes under Full proof where they are not. Results this unit does not prove are marked cited, with a reference. Terms with a dotted underline are glossary entries — click one, or press G. N opens these notes; R stacks the deck into one readable page.

11 · overview

One question, asked of a curve

the object

A functional $J$ eats a whole function and returns one number: $J[y] = \int_a^b F(x, y(x), y'(x))\,\mathrm{d}x$.

the derivative

Perturb $y$ to $y + \varepsilon\eta$ and differentiate in $\varepsilon$ at $0$. That number, the first variation, is linear in $\eta$ — a directional derivative on a space of curves.

the equation

"Zero for every $\eta$" collapses, by one lemma, into a single ODE: the Euler–Lagrange equation $\ \dfrac{\mathrm d}{\mathrm dx}F_{y'} = F_y$.

the running example

A bead slides on a frictionless wire from $A = (0,0)$ to $B = (2\ \mathrm m, 1\ \mathrm m)$, $y$ measured downwards, from rest, $g = 9.81\ \mathrm{m\,s^{-2}}$. Which wire is quickest?

Straight: $1.00964$ s. Galileo's circular arc: $0.83031$ s. The winner is a cycloid, $0.80556$ s, reached in four steps:

1. Energy gives the speed, $v = \sqrt{2gy}$, so the time $\int\mathrm ds/v$ is a functional of the wire.

2. Its integrand has no $x$ in it, so the equation has a first integral: $y(1+y'^2) = 2a$.

3. Substituting $y = a(1-\cos\theta)$ integrates that: $x = a(\theta - \sin\theta)$.

4. $B$ fixes $a$ and the end angle; the second variation confirms a minimum.

The logical order matters and is easy to lose. The Euler–Lagrange equation is a necessary condition for a minimum, in exactly the way $f'(x) = 0$ is necessary for a minimum of a function of one variable. It does not by itself say a minimum exists, and it does not distinguish minima from maxima or from saddles. Sections 5 and 6 are about those two gaps, and they are not decoration: unit 26 will need to know that a free particle's worldline in Minkowski space makes proper time a maximum, not a minimum, and that the geodesics of unit 32 stop being minimal past a conjugate point.

Getting around. A dotted underline is a glossary term: click it, or press G for the index. N opens these notes, where the longer proofs live. A solid underline, like the running example, shows a piece of the bead problem on hover.

The four steps, and where each is done. Step 1 is The bead problem, written down (the wire's reaction does no work, so speed depends on depth alone). Step 2 is the Beltrami identity — The Beltrami identity: when $x$ is absent proves it for any integrand without $x$, and Beltrami on the bead evaluates it; the constant is named $1/\sqrt{4ga}$, so $a$ is a length. Step 3 is Worked: integrating the first integral, a half-angle substitution. Step 4 is split: The constants, and the race fixes $\theta_B$ and $a$ from $X/Y = 2$, and section 5 — ending at Worked: the cycloid has no conjugate point — checks that the stationary curve is really a minimum, which is not automatic.

Three numbers on this slide are computed, not quoted: $1.00964$ s from the closed form $\sqrt{2(X^2+Y^2)/(gY)}$, $0.83031$ s by quadrature along the circular arc, and $0.80556$ s from $\theta_B\sqrt{a/g}$. Section 3 does all three.

section 1

Functionals, and what it means to vary one

Three old problems with the same shape, and the derivative that fits them.

11 · functionals

Three problems with one shape

Fermat, 1662

Light travels between two points along the path of least time. With refractive index $n = c/v$,

$$\text{minimise}\ \int n\,\mathrm{d}s .$$

Snell's law falls out in one line, next slide.

Bernoulli, 1696

The brachistochrone: the wire down which a bead, released from rest, descends between two points in least time,

$$\text{minimise}\ \int \frac{\mathrm{d}s}{v} .$$

Six months' grace, then Newton solved it overnight.

the chain

A hanging chain minimises its potential energy,

$$\text{minimise}\ \int \rho g\, y\,\mathrm{d}s$$

with the length $\int\mathrm{d}s$ held fixed.

Galileo said parabola. He was wrong; section 4 says what it is.

the shape

In every one: a number attached to a whole curve, an integral of a local quantity along it, and an instruction to make it as small as possible. What differential equation does that instruction force?

The history, briefly. Johann Bernoulli posed the brachistochrone in Acta Eruditorum in June 1696 as a challenge to "the sharpest mathematicians of all the world", allowing six months. Besides his own, five solutions arrived: his brother Jacob's, Leibniz's, l'Hôpital's, Tschirnhaus's and an anonymous one from England, which Bernoulli is said to have recognised at once — tanquam ex ungue leonem, the lion is known by its claw. Newton had received the problem on 29 January 1697 and sent his answer the next day.

Johann's own solution was a trick rather than a method: he sliced the region into thin horizontal layers of constant speed, applied Snell's law at each interface, and passed to the limit. It gives the right answer and generalises to nothing. Jacob's solution was clumsier and far more important, because it varied the curve locally and is the ancestor of the argument in section 2. The brothers' rivalry over it was poisonous and lasted the rest of their lives.

Euler systematised the subject in 1744, by discretising the curve into $n$ points, applying ordinary calculus, and letting $n \to \infty$. Lagrange, aged 19, wrote to him in 1755 with the $\delta$-calculus used here — variations of the whole function at once, no discretisation. Euler adopted it, named the subject the calculus of variations, and let the boy publish first.

The third problem, the catenaria, was proposed by Jacob Bernoulli in 1690 and solved in 1691 by Huygens, Leibniz and Johann Bernoulli. Galileo had asserted a parabola in 1638; Huygens had shown as a teenager in 1646 that it is not one.

11 · functionals · example

Worked: Snell's law, from Fermat

minimise over one number

Two media meet on the $x$-axis: speed $v_1$ above, $v_2$ below. Light goes from $P = (0, h_1)$ to $Q = (d, -h_2)$. Inside each medium the quickest route is straight, so the only freedom is the crossing point $(x, 0)$:

$$T(x) = \frac{\sqrt{x^2+h_1^2}}{v_1} + \frac{\sqrt{(d-x)^2+h_2^2}}{v_2}$$

Step 1. Differentiate: $\ T'(x) = \dfrac{x}{v_1\sqrt{x^2+h_1^2}} - \dfrac{d-x}{v_2\sqrt{(d-x)^2+h_2^2}}$.

Step 2. Read the two fractions as sines of the angles from the normal: $\sin\theta_1 = x/\sqrt{x^2+h_1^2}$, $\sin\theta_2 = (d-x)/\sqrt{(d-x)^2+h_2^2}$.

Step 3. $T'(x) = 0$ is therefore Snell's law:

$$\frac{\sin\theta_1}{v_1} = \frac{\sin\theta_2}{v_2}, \qquad n_1\sin\theta_1 = n_2\sin\theta_2 .$$

The only freedom is where the ray crosses. It bends towards the normal in the slower medium, spending less of its path there than the straight segment (dotted) does.

Reading the figure. The two sines in Step 2 are read straight off the two right triangles: the one above the interface has legs $x$ and $h_1$, the one below $d - x$ and $h_2$, and in each the angle at the crossing point between the ray and the vertical normal is the $\theta$ drawn. The straight segment $PQ$ is the shortest path but spends more of its length in the slow medium; the true ray trades a little extra length above for less below.

Total internal reflection is the failure of this construction to have a solution: going from water to air, $\sin\theta_2 = 1.333\sin\theta_1$ exceeds $1$ once $\theta_1 > \arcsin(1/1.333) = 48.61^\circ$, and no crossing point solves the equation. The ray does not cross.

Johann Bernoulli's original brachistochrone proof was exactly this slide, iterated: stack thin horizontal layers with $v = \sqrt{2gy}$ constant in each, apply Snell at every interface to get $\sin\theta/v$ constant, and let the layers thin. Section 3 recovers $\sin\theta = \text{const}\times\sqrt{y}$ as the Beltrami identity, which is the same statement with the scaffolding removed.

11 · functionals · example

Snell's law: least, not just stationary

checks

Minimum, not just stationary. Each term of $T$ is a composition of the convex $u \mapsto \sqrt{u^2+h^2}$ with an affine map, hence convex, and strictly so; the sum is strictly convex, so the stationary point is the unique global minimum.

Limit. $v_1 = v_2$ gives $\theta_1 = \theta_2$: no bend, a straight line, as it must be.

A number. Air to water, $n_1 = 1$, $n_2 = 1.333$, $\theta_1 = 45^\circ$: $\sin\theta_2 = 0.70711/1.333 = 0.53046$, so $\theta_2 = 32.04^\circ$.

One unknown number, so ordinary calculus sufficed. Curve the interface, or let $n$ vary continuously, and the unknown becomes a function — which is the whole of this unit.

That example with $h_1 = h_2 = h$, so $d = h(1 + \tan 32.04^\circ) = 1.626\,h$. Two convex bowls (dashed) add to one, and a convex function has at most one stationary point: the minimum, at $x = h$, where $\theta_1 = 45^\circ$.

Why "least" is safe here and not in general. Convexity is what makes the word honest. The brachistochrone's integrand is not convex in $(y, y')$, so nothing this cheap is available there, and section 5 has to work for the same conclusion. This is the first appearance of a pattern that runs to the end of the course: a stationarity condition is easy, and deciding what kind of stationary point you have is not.

Fermat's principle is not exactly "least time". The correct statement is that the optical path length is stationary. A ray from one focus of an elliptical mirror to the other has the same optical length for every reflection point — stationary and not isolated at all — and a ray reflected inside a suitably curved mirror can be a local maximum. Fermat asserted least time in 1662 and was attacked by the Cartesians for smuggling teleology into physics; the stationary form removes the objection, and unit 15 removes it entirely by showing where the principle comes from.

How the curve was computed. $T(x) = \sqrt{x^2+1} + 1.333\sqrt{(d-x)^2+1}$ in units of $h/v_1$, with $d = 1 + \tan\theta_2 = 1.62574$ so that the stationary point sits at $x = 1$, $\theta_1 = 45^\circ$; its value there is $1.41421 + 1.333\times1.17964 = 2.98667$.

11 · functionals

The bead problem, written down

Take $A$ at the origin and measure $y$ downwards, so the bead descends as $y$ grows. Along the wire:

Speed. The frictionless wire's reaction is normal to the motion, so does no work (unit 04). From rest, $\tfrac12 mv^2 = mgy$:

$$v = \sqrt{2gy} .$$

Arc length. For a graph $y(x)$, unit 03 gives $\mathrm{d}s = \sqrt{1+y'^2}\,\mathrm dx$.

Time. $\mathrm dt = \mathrm ds/v$, and integrate:

the functional $$T[y] = \int_0^{X}\sqrt{\frac{1+y'(x)^2}{2g\,y(x)}}\ \mathrm{d}x$$

over curves with $y(0) = 0$, $y(X) = Y$, and $y > 0$ on $(0, X]$.

Schematic. Whatever the wire, the bead at depth $y$ has speed $\sqrt{2gy}$; the triangle turns a step $\mathrm dx$ into the length $\mathrm ds$ actually travelled.

What to look for in the figure. Two separate facts, multiplied together. The speed is a function of depth alone — the dashed vertical — and not of the route taken to that depth. The length element comes from the small triangle by Pythagoras, $\mathrm ds^2 = \mathrm dx^2 + (y'\,\mathrm dx)^2$. The time for the step is their ratio, and nothing else about the wire enters.

Why the constraint force does no work. A bead threaded on a smooth wire feels a reaction $\vec N$ perpendicular to the wire, so $\vec N\cdot\vec v = 0$ pointwise. Unit 04's work–energy theorem then gives $\tfrac12 mv^2 = \int \vec F\cdot\mathrm d\vec r$ with only gravity contributing, and gravity is conservative with potential energy $-mgy$ for $y$ measured downwards. Hence $v = \sqrt{2gy}$ along any wire whatsoever, which is what makes the problem tractable: the speed depends on the height alone and not on the route taken to it.

A restriction that is real. Writing the curve as a graph $y(x)$ assumes it never doubles back in $x$. The true minimiser does not double back, but that is a fact about the answer, not an assumption one is entitled to. Problem P1 part 4 and the geodesic problems make the point again: an extremal found among graphs is an extremal among graphs.

11 · functionals

The bead's functional: no mass, and a singular start

dimensions

$[y'] = 1$, $[gy] = \mathrm{m^2\,s^{-2}}$, so the square root is $\mathrm{s\,m^{-1}}$ and $\times\,\mathrm dx$ gives seconds ✓.

two things to notice

The mass has vanished: every bead takes the same time.

At $x = 0$, $y = 0$ and the integrand blows up. Whether the improper integral converges depends on how the wire leaves $A$: steeply is fine, flat is fatal.

Three wires to $B$ and the rate at which each integrand blows up at $A$. A wire leaving flat never gets the bead started — the first hint that the answer sets off steeply down.

The improper integral, carefully. Near $x = 0$ a curve leaving the origin with finite slope $m > 0$ has $y(x) = mx + O(x^2)$ and $y' \to m$, so the integrand behaves like $\sqrt{(1+m^2)/(2gm)}\,x^{-1/2}$, whose integral converges. The cycloid leaves the origin vertically — infinite slope — and still converges, though its integrand is the more singular of the two: there $y \sim (9a/2)^{1/3}x^{2/3}$ and the integrand goes like $x^{-2/3}$. A curve leaving horizontally can fail altogether: for $y \approx kx^2$ the integrand behaves like $x^{-1}/\sqrt{2gk}$, whose integral diverges — the bead never starts. The dashed parabola in the figure is $k = 1/4$, chosen to pass through $B$. This is the first hint that the answer will want to set off steeply down.

Mass-independence is unit 00's mechanical similarity in a new dress. Scaling the whole picture by $\lambda$ and times by $\sqrt\lambda$ leaves the problem invariant; $m$ never enters because it divides out of $m\ddot{\vec r} = m\vec g + \vec N$ only after $\vec N$ has been eliminated by the no-work argument — which is exactly what the energy route does.

11 · functionals

The space the unknown lives in

the admissible set

$$\mathcal A = \{\,y \in C^1[a,b] \;:\; y(a) = \alpha,\ y(b) = \beta\,\}$$

Not a vector space: the sum of two members takes the value $2\alpha$ at $a$. It is an affine subspace — a fixed $y_0 \in \mathcal A$ plus the vector space

$$C^1_0 = \{\,\eta\in C^1[a,b] : \eta(a) = \eta(b) = 0\,\} .$$

admissible variation

An element $\eta \in C^1_0$. Then $y + \varepsilon\eta \in \mathcal A$ for every real $\varepsilon$: the whole line through $y$ in direction $\eta$ stays admissible. This is why the boundary conditions never interfere with the differentiation.

The variation is pinned at both ends. Every curve on the dashed family is admissible, so $\varepsilon\mapsto J[y+\varepsilon\eta]$ is an ordinary function of one real variable.

Why $C^1$ and not $C^2$. The functional needs $y'$ and nothing more, so $C^1$ is the natural class: every competitor must have a slope for $T[y]$ to mean anything. The usual form of the Euler–Lagrange equation contains $y''$, and it is tempting to demand $C^2$ so that it can be written down — but that would assume what one wants to prove. Section 2's du Bois-Reymond lemma works with $y \in C^1$, and then a regularity theorem hands the second derivative back wherever $F_{y'y'} \ne 0$ (problem P3). Admitting corners — piecewise $C^1$ curves — is a real extension that changes the answers to some problems; the Weierstrass–Erdmann corner conditions govern it, and they are cited here, not proved: Gelfand and Fomin, Calculus of Variations, §15.

The infinite-dimensionality is the whole difficulty. Unit 04 minimised over $\mathbb R^n$, where closed bounded sets are compact and a continuous function on one attains its minimum. $C^1[a,b]$ has no such theorem: its closed unit ball is not compact in any of the natural norms. Section 6 is about what goes wrong when one forgets that.

Notation. $\eta$ for the variation, $\varepsilon$ for its size, $\delta y = \varepsilon\eta$ for the pair together. Lagrange's $\delta$ is a genuine directional derivative and not an infinitesimal; the notes keep $\varepsilon$ explicit so that every limit taken is one the learner has already seen.

11 · functionals

The first variation is a directional derivative

definition

The first variation of $J$ at $y$ in the direction $\eta$ is the Gateaux derivative

$$\delta J[y;\eta] \;=\; \left.\frac{\mathrm d}{\mathrm d\varepsilon}\right|_{\varepsilon = 0} J[y+\varepsilon\eta] \;=\; \lim_{\varepsilon\to0}\frac{J[y+\varepsilon\eta]-J[y]}{\varepsilon},$$

when the limit exists. Nothing new is being defined: $\varphi(\varepsilon) = J[y+\varepsilon\eta]$ is a function $\mathbb R\to\mathbb R$, and this is $\varphi'(0)$.

theorem · the formula

If $F$ is $C^2$ on a neighbourhood of the graph data, $y \in C^1[a,b]$ and $\eta\in C^1_0$, then $\delta J[y;\eta]$ exists and

$$\delta J[y;\eta] = \int_a^b\Big(F_y\big(x,y,y'\big)\,\eta(x) + F_{y'}\big(x,y,y'\big)\,\eta'(x)\Big)\,\mathrm{d}x ,$$

and it is linear in $\eta$.

differentiation under the integral sign — full proof in the notes (N)

Necessary condition, exactly as in one variable: if $y$ minimises $J$ over $\mathcal A$, then $\varphi(\varepsilon) = J[y+\varepsilon\eta]$ has a minimum at $\varepsilon=0$, so $\varphi'(0) = \delta J[y;\eta] = 0$ — for every $\eta$.

Full proof

Write $\varphi(\varepsilon) = \int_a^b G(x,\varepsilon)\,\mathrm dx$ with $G(x,\varepsilon) = F\big(x,\,y(x)+\varepsilon\eta(x),\,y'(x)+\varepsilon\eta'(x)\big)$. For fixed $x$, unit 03's chain rule gives $\partial_\varepsilon G = F_y\,\eta + F_{y'}\,\eta'$, evaluated at the shifted arguments. We must justify exchanging $\partial_\varepsilon$ with $\int$.

Fix $|\varepsilon| \le 1$. The set $K = \{(x, y(x)+t\eta(x), y'(x)+t\eta'(x)) : x\in[a,b],\ |t|\le 1\}$ is the continuous image of the compact $[a,b]\times[-1,1]$, hence compact, and $\partial_\varepsilon G$ is continuous on it, hence uniformly continuous and bounded there. For $\varepsilon \ne 0$, the mean value theorem in $\varepsilon$ gives, for each $x$, some $\tau(x)$ between $0$ and $\varepsilon$ with

$$\frac{G(x,\varepsilon)-G(x,0)}{\varepsilon} = \partial_\varepsilon G(x,\tau(x)).$$

By uniform continuity, given $\sigma>0$ there is $\delta>0$ with $|\partial_\varepsilon G(x,\tau) - \partial_\varepsilon G(x,0)| < \sigma$ whenever $|\tau| < \delta$, uniformly in $x$. So for $|\varepsilon| < \delta$ the difference quotient of $\varphi$ differs from $\int_a^b \partial_\varepsilon G(x,0)\, \mathrm dx$ by at most $\sigma(b-a)$. Letting $\sigma\to0$ gives the claim. The same argument with $\partial^2_\varepsilon G$ gives $\varphi''(0)$, which is section 5's second variation.

Linearity in $\eta$ is visible in the formula: $\eta$ and $\eta'$ each appear once, multiplied by something that does not involve them. It is the reason the condition "$\delta J = 0$ for all $\eta$" can be turned into a pointwise statement at all — it says a linear functional on $C^1_0$ vanishes identically.

Gateaux versus Fréchet. The Gateaux derivative asks only for directional derivatives; the Fréchet derivative asks for a single linear map approximating $J$ uniformly over directions. Gateaux is weaker, is all that is needed for a necessary condition, and — unlike Fréchet — costs nothing to define. The distinction bites in section 5, where the second-order behaviour depends on which norm is used to measure "small $\eta$".

The brachistochrone's improper integrand is not covered by the compactness argument above, since $F$ blows up at $y = 0$. The repair is routine and is the reason the endpoint is treated separately: work on $[\epsilon, X]$, where everything is uniform, and check that the contribution of $[0,\epsilon]$ tends to zero as $\epsilon\to0$, which the $x^{-1/2}$ bound on the bead slide supplies.

11 · functionals

Which minimum? Two norms, two answers

two notions of nearby

$\lVert \eta\rVert_0 = \max|\eta|$ and $\lVert\eta\rVert_1 = \max|\eta| + \max|\eta'|$.

$y$ is a weak local minimum if $J[y]\le J[\tilde y]$ for all admissible $\tilde y$ with $\lVert\tilde y - y\rVert_1$ small; a strong local minimum if the same holds with $\lVert\cdot\rVert_0$ small.

Strong is the stronger property: there are more competitors, because a curve may be close in position while wildly different in slope.

"least action" is a misnomer

$\delta J[y;\eta] = 0$ for all $\eta$ makes $y$ stationary — an extremal. It does not make $J[y]$ least, or even locally least.

Exactly as $f'(x) = 0$ admits maxima and inflections. The reflex "extremal, therefore minimum" is the single most common error in this subject, and unit 26 will punish it: a free particle's worldline maximises proper time.

Section 5 supplies the tests — the Legendre condition and Jacobi's conjugate points — that tell the cases apart, and applies them to the cycloid.

Why two norms are needed is the next slide, with a picture: a curve can stay within any $\delta$ of $y$ in position while its slope is nowhere near $y'$, and some functionals notice only the slope.

Where the phrase came from. Maupertuis announced a "principle of least action" in 1744 with theological framing — nature is economical because its author is. Euler had the correct mathematical statement the same year and made no such claim. Jacobi, who introduced conjugate points in 1837, pointed out in his lectures on dynamics of 1842–43 that the action is stationary and often not least, which is where conjugate points enter mechanics. The course says stationary action throughout, and unit 12 onwards keeps to it.

Neither norm makes $C^1_0$ compact. The functions $\eta_n(x) = n^{-1}\sin(n\pi x)$ are bounded in $\lVert\cdot\rVert_1$ with no convergent subsequence in it. That is the obstruction section 6 has to work around.

11 · functionals

Close in position, far in slope

weak, and not strong

$J[y] = \displaystyle\int_0^1 y'^2(1-y')\,\mathrm dx$, $\ y(0) = y(1) = 0$.

Weak minimum at $y\equiv0$. If $\lVert\tilde y\rVert_1 < 1$ then $|\tilde y'| < 1$, so the integrand is $\ge 0 = J[0]$.

Not strong. A sawtooth with slope $+2$ on a fraction $p$ of each tooth and $-2p/(1-p)$ on the rest gives

$$J = -4p + \frac{4p^2(1+p)}{(1-p)^2}$$

however many teeth: $-0.34568$ at $p = 0.1$. More, smaller teeth put it within any $\delta$ of $0$.

Drawn with $p = 1/4$, where $J = -0.444$. In $\lVert\cdot\rVert_0$ the sawtooth is as close to $0$ as the bump is; in $\lVert\cdot\rVert_1$ it is at distance more than $2$, so a weak minimum never has to compete with it.

The computation. On the rising part of each tooth $y' = 2$, so $y'^2(1-y') = 4\cdot(1-2) = -4$, over a fraction $p$ of the interval: $-4p$. On the falling part $y' = -2p/(1-p)$ — chosen so that each tooth climbs and descends the same height, and the teeth close up — and $y'^2(1-y') = \frac{4p^2}{(1-p)^2}\cdot\frac{1+p}{1-p}$, over a fraction $1 - p$: $4p^2(1+p)/(1-p)^2$. The total does not depend on the number of teeth, while the height of each tooth is $2p$ times its width, so more teeth make the curve uniformly as close to $0$ as one likes. At $p = 0.1$, $-0.4 + 0.04\times1.1/0.81 = -0.34568$; at $p = 1/4$, $-1 + 0.25\times1.25/0.5625 = -0.44444$.

For contrast, $\int_0^1 y'^2(1-y')^2\,\mathrm dx$ has a non-negative integrand, so $y\equiv0$, with $J = 0$, is a global minimum in the strong sense as well: the sawtooth only helps when steep slopes can make the integrand negative. This is the standard demonstration that the Legendre condition of section 5 is a statement about weak minima.

section 2

From "for every $\eta$" to one equation

Two lemmas, both proved by building a function that is zero except where you want it.

11 · the lemmas

The tool: a bump

bump function

Given $[c,d]\subset(a,b)$, define

$$\eta_{c,d}(x) = \begin{cases}\big[(x-c)(d-x)\big]^{3}, & c\le x\le d,\\[2pt] 0 & \text{otherwise.}\end{cases}$$

it is $C^2$ on $[a,b]$ and positive exactly on $(c,d)$

On $(c,d)$ both factors are positive, so $\eta > 0$; outside $[c,d]$ it is $0$. At $x = c$ the polynomial has a triple root, so it and its first two derivatives vanish there — the one-sided derivatives from inside and outside agree, and they are continuous. Same at $d$. Hence $\eta_{c,d} \in C^2_0$.

A cube, not a square, precisely so that two derivatives vanish at the ends. For $C^k$ use the $(k+1)$st power; for $C^\infty$ use $\exp\!\big({-}1/((x-c)(d-x))\big)$ on $(c,d)$ and $0$ outside.

Zero outside $[c,d]$, strictly positive inside, and pinned at $a$ and $b$ automatically. The whole of section 2 is this picture used twice.

Why a bump is the right tool. The statement to be defeated is "an integral against every test function vanishes". A bump concentrated where the unknown is non-zero converts that global statement into a sign contradiction at one point. It is the same move as proving a continuous function is zero by testing it against approximate delta functions — and unit 20 makes that literal, defining distributions as exactly the things that can be tested this way.

The $C^\infty$ bump, since it is cited. The function $f(t) = e^{-1/t}$ for $t>0$ and $0$ for $t\le0$ is $C^\infty$: by induction, $f^{(n)}(t) = p_n(1/t)e^{-1/t}$ for a polynomial $p_n$ when $t > 0$, and every such expression tends to $0$ as $t\downarrow0$ because $e^{-1/t}$ beats any power of $1/t$. Composing, $\eta(x) = f\big((x-c)(d-x)\big)$ is $C^\infty$, positive exactly on $(c,d)$. The polynomial bump is used on the slide because its smoothness needs no induction, and $C^2$ is all the lemmas below consume.

A caution about the interval. The lemma below needs $[c,d]$ strictly inside $(a,b)$ so that $\eta$ vanishes at the endpoints. When the endpoints themselves are in question — the free-boundary problem of section 4 — one deliberately does not restrict to such $\eta$, and the surviving boundary term is the whole content of the natural boundary condition.

11 · the lemmas

The fundamental lemma of the calculus of variations

theorem · the fundamental lemma

Let $M$ be continuous on $[a,b]$. If $\displaystyle\int_a^b M(x)\,\eta(x)\,\mathrm dx = 0$ for every $\eta\in C^2_0[a,b]$ — so in particular if it holds for every $\eta \in C^1_0$ — then $M \equiv 0$ on $[a,b]$.

proof

Suppose not: $M(x_0)\ne0$ for some $x_0\in[a,b]$, and without loss of generality $M(x_0) > 0$ (else replace $M$ by $-M$). By continuity $M > M(x_0)/2$ on some interval around $x_0$, and that interval contains some $[c,d]\subset(a,b)$ with $c < d$. Take $\eta = \eta_{c,d}$, the bump. Then

$$\int_a^b M\eta = \int_c^d M\eta \;\ge\; \frac{M(x_0)}{2}\int_c^d\eta \;>\; 0,$$

since $\eta > 0$ on $(c,d)$. That contradicts the hypothesis. ∎

The proof in one picture. The bump switches the integral off except where $M$ is safely positive, so the shaded integrand $M\eta$ cannot integrate to zero.

Continuity is doing real work. An $M$ that is non-zero at one point only integrates to zero against everything. The lemma says a continuous function is determined by its integrals against bumps — and the figure is why: continuity is what supplies the interval $[c,d]$ on which $M$ stays above $M(x_0)/2$.

The endpoint case. If the only point with $M \ne 0$ were $x_0 = a$, continuity still supplies an interval $[a, a+\rho)$ on which $M > M(a)/2$, and any $[c,d]\subset(a, a+\rho)$ works. So the conclusion holds on the closed interval.

What "without loss of generality" costs here. Nothing: the hypothesis is symmetric under $M\mapsto -M$, because $\int(-M)\eta = -\int M\eta = 0$ too.

Sharpenings that matter later. The same proof works if $\eta$ ranges only over $C^\infty_0$, which is the version unit 20 uses, and if $M$ is merely locally integrable the conclusion becomes "$M = 0$ almost everywhere" — the du Bois-Reymond lemma of measure theory, which is cited, not proved here. Neither strengthening is needed for this unit.

Name. Lagrange used the argument in 1755 without isolating it; the careful statement is due to du Bois-Reymond (1879), who was also the first to worry about whether extremals had to be twice differentiable. It is sometimes called the fundamental lemma, sometimes Lagrange's lemma.

11 · the lemmas

The du Bois-Reymond lemma

theorem · du Bois-Reymond lemma

Let $N$ be continuous on $[a,b]$. If $\displaystyle\int_a^b N(x)\,\eta'(x)\,\mathrm dx = 0$ for every $\eta\in C^1_0[a,b]$, then $N$ is constant.

proof

Let $\bar N = \frac1{b-a}\int_a^b N$ be the mean, and set $\eta(x) = \int_a^x\big(N(t)-\bar N\big)\,\mathrm dt$. Then $\eta\in C^1$, $\eta(a) = 0$, and $\eta(b) = \int_a^b N - (b-a)\bar N = 0$, so $\eta\in C^1_0$ — it is an admissible variation. Its derivative is $\eta' = N - \bar N$. Now

$$0 = \int_a^b N\eta'\,\mathrm dx = \int_a^b N\big(N-\bar N\big)\,\mathrm dx = \int_a^b\big(N-\bar N\big)^2\mathrm dx,$$

the last step because $\int_a^b \bar N(N-\bar N) = \bar N\cdot 0 = 0$. A continuous non-negative integrand with zero integral vanishes identically, so $N \equiv \bar N$. ∎

No bump was needed here — the test function was built out of $N$ itself. Combined with one integration by parts, it is the whole proof of the Euler–Lagrange equation, on the next slide.

Why this lemma and not just the first one. The first variation contains $\eta'$, and removing it by integrating $F_{y'}\eta'$ by parts requires $\frac{\mathrm d}{\mathrm dx}F_{y'}$ to exist — which presumes $y''$ exists, which is what one is trying to find out. Du Bois-Reymond's lemma takes the first variation as it stands and yields a statement with no second derivatives in it. That statement, the integrated form on the next slide, is the honest Euler–Lagrange equation for a $C^1$ extremal.

The combined form, which is what an extremal actually satisfies: if $M, N$ are continuous and $\int_a^b (M\eta + N\eta')\,\mathrm dx = 0$ for every $\eta\in C^1_0$, then $N$ is $C^1$ and $N' = M$. Proof: let $P(x) = \int_a^x M$, so $P' = M$ and $\int M\eta = [P\eta]_a^b - \int P\eta' = -\int P\eta'$ since $\eta$ vanishes at both ends. The hypothesis becomes $\int (N - P)\eta' = 0$ for all such $\eta$, so by the lemma $N - P$ is a constant $k$; hence $N = P + k$ is differentiable with $N' = M$. Problem P3 asks for this.

A continuous non-negative function with zero integral is zero: if $h(x_0) > 0$ then $h > h(x_0)/2$ on some interval of length $\ell > 0$, giving $\int h \ge \ell h(x_0)/2 > 0$. Used twice in this section and worth having reflexively.

11 · the equation

The Euler–Lagrange equation

theorem

Let $F$ be $C^2$ and let $y\in C^1[a,b]$ be an extremal of $J[y] = \int_a^b F(x,y,y')\,\mathrm dx$ on $\mathcal A$. Then $x\mapsto F_{y'}(x,y,y')$ is $C^1$ and

integrated (du Bois-Reymond) form

$$F_{y'}(x,y,y') = \int_a^x F_y\,\mathrm dt + \text{const}$$

differentiated form, if $y\in C^2$

$$\frac{\mathrm d}{\mathrm dx}F_{y'}(x,y,y') = F_y(x,y,y')$$

proof

$\delta J[y;\eta] = \int_a^b(F_y\eta + F_{y'}\eta') \,\mathrm dx= 0$ for every $\eta\in C^1_0$. Put $W(x) = \int_a^x F_y\,\mathrm dt$, so $W' = F_y$; integrating by parts, $\int_a^b F_y\eta = [W\eta]_a^b - \int_a^b W\eta' = -\int_a^b W\eta'$, since $\eta$ vanishes at both ends. So $\int_a^b(F_{y'} - W)\,\eta'\,\mathrm dx = 0$ for every $\eta$, and the du Bois-Reymond lemma makes $F_{y'} - W$ a constant: that is the left form. Its right side is $C^1$, so $F_{y'}$ is too, and differentiating gives the right form. ∎

Expanded by the chain rule, the right-hand form is $F_{y'y'}\,y'' + F_{yy'}\,y' + F_{xy'} = F_y$: a second-order ODE, nonlinear in general, and degenerate exactly where $F_{y'y'} = 0$.

Read the equation before solving any. It is second order, so its general solution carries two constants, which is exactly the right count to meet two boundary conditions $y(a)=\alpha$, $y(b)=\beta$. That is a boundary-value problem, not the initial-value problem of units 00 and 01, and the difference is not cosmetic: a boundary-value problem may have no solution, one, or many. The equation $y'' + y = 0$ with $y(0)=0$, $y(\pi)=0$ has infinitely many solutions $y = c\sin x$; with $y(\pi) = 1$ it has none. (Proof of "none": integrating by parts twice, any $C^2$ solution has $0 = \int_0^\pi(y''+y)\sin x\,\mathrm dx = \big[y'\sin x - y\cos x\big]_0^\pi = y(\pi) + y(0)$, so $y(\pi) = -y(0) = 0$.) That failure is conjugate points, met again in section 5 with a reason attached.

Why $F_{y'y'}$ decides everything. The expanded form can be solved for $y''$ only where $F_{y'y'} \ne 0$. Where it vanishes the equation is not an ODE for $y''$ at all but a constraint — which is what happens to the relativistic free particle in unit 27 and, in a much more serious way, to the Einstein–Hilbert action's lapse and shift in unit 35. The same quantity is the Legendre condition's subject in section 5, and unit 14 shows that its non-vanishing is precisely what makes the Legendre transform to a Hamiltonian possible. One number, three jobs.

Regularity. The theorem as stated assumes only $y \in C^1$ and concludes that $F_{y'}$ is $C^1$. If in addition $F_{y'y'}\ne0$ along the extremal, the implicit function theorem of unit 04 — applied to $p\mapsto F_{y'}(x,y,p)$ — gives $y' $ as a $C^1$ function of $x$, so $y \in C^2$ after all. This is Hilbert's differentiability theorem; problem P3 walks through it.

Terminology. $F$ is called the integrand here and the Lagrangian from unit 12 onwards, where it becomes $L = T - V$ and $x$ becomes time. The equation is due to Euler (1744) in the discretised form and to Lagrange (1755) in this one.

11 · the equation · example

Worked: the shortest path, and what the derivation assumed

arc length in the plane

$L[y] = \displaystyle\int_a^b\sqrt{1+y'^2}\,\mathrm dx$, so $F = \sqrt{1+y'^2}$.

Step 1. $F_y = 0$ — the integrand has no $y$ in it.

Step 2. $F_{y'} = \dfrac{y'}{\sqrt{1+y'^2}}$.

Step 3. Euler–Lagrange therefore reads $\dfrac{\mathrm d}{\mathrm dx}\dfrac{y'}{\sqrt{1+y'^2}} = 0$, so $\dfrac{y'}{\sqrt{1+y'^2}} = C$.

Step 4. Then $y'^2 = C^2(1+y'^2)$, so $y'^2 = C^2/(1-C^2)$, a constant: $y = mx + k$, a straight line, $m$ and $k$ fixed by the endpoints.

checks

Dimensions. $F$ is dimensionless, $L$ has the dimension of $x$ ✓.

Sign. $|C| < 1$ is forced, since $|y'|/\sqrt{1+y'^2} < 1$ always; $C \to \pm1$ is the vertical line the graph formulation cannot represent.

Legendre. $F_{y'y'} = (1+y'^2)^{-3/2} > 0$ everywhere, so section 5's necessary condition for a minimum passes.

what was assumed

That competitors are graphs over $[a,b]$. So: among curves crossing each vertical line once, the segment is shortest. All curves need another argument (notes).

When $F_y = 0$, $F_{y'}$ is a first integral, conserved along every extremal — unit 13's Noether theorem for translations in $y$.

The pattern, and it is the one used for the rest of the course. When the integrand does not contain $y$, the quantity $F_{y'}$ is constant along extremals; when it does not contain $x$, the next slide's Beltrami combination is constant. Each is a symmetry giving a conserved quantity, and each drops the order of the ODE by one. Unit 08 did the same thing by hand for central forces — energy and angular momentum taking a six-dimensional system down to a quadrature — and unit 13 will prove that all of these are the same theorem.

The vertical-line case is not a technicality. The cycloid of section 3 leaves the origin vertically, and the graph formulation survives only because it is vertical at a single endpoint rather than over an interval. Problem P1 part 4 asks for a functional whose true minimiser is not a graph at all, so that the Euler–Lagrange equation finds nothing.

An honest proof that the segment is shortest among all curves, for comparison, and it uses none of this machinery: for a piecewise $C^1$ curve $\vec\gamma$ from $\vec p$ to $\vec q$, let $\hat u = (\vec q-\vec p)/\lVert\vec q-\vec p\rVert$; then $\lVert\vec q - \vec p\rVert = \int \dot{\vec\gamma}\cdot\hat u \,\mathrm dt \le \int\lVert\dot{\vec\gamma}\rVert\,\mathrm dt = L$, by Cauchy–Schwarz, with equality only when $\dot{\vec\gamma}$ is a non-negative multiple of $\hat u$ throughout. The variational route earns its keep when no such trick exists — which is every other problem in this deck.

11 · the equation

The Beltrami identity: when $x$ is absent

theorem · Beltrami identity

Along any $C^2$ extremal, $\ \dfrac{\mathrm d}{\mathrm dx}\Big(F - y'F_{y'}\Big) = F_x$. In particular, if $F$ does not depend explicitly on $x$, then

$$F - y'\,F_{y'} = \text{constant along the extremal.}$$

proof · one line

By the chain rule, $\dfrac{\mathrm dF}{\mathrm dx} = F_x + F_y y' + F_{y'}y''$, and $\dfrac{\mathrm d}{\mathrm dx}\big(y'F_{y'}\big) = y''F_{y'} + y'\dfrac{\mathrm d}{\mathrm dx}F_{y'}$. Subtract:

$$\frac{\mathrm d}{\mathrm dx}\big(F - y'F_{y'}\big) = F_x + y'\underbrace{\Big(F_y - \tfrac{\mathrm d}{\mathrm dx}F_{y'}\Big)}_{=\,0\ \text{by Euler–Lagrange}} .$$

The $F_{y'}y''$ terms cancel and the bracket vanishes, leaving $F_x$. ∎

Why it is worth having. It replaces a second-order ODE by a first-order one, which unit 00's separation of variables can then attack. The two integration constants become one constant here and one from the quadrature.

What it will become. Put $p = F_{y'}$ and $H = y'p - F$. The identity says $H$ is conserved when $F$ has no explicit $x$. It obeys $\mathrm dH/\mathrm dx = -F_x$. With $x$ as time and $F = T-V$ this is energy conservation, and $H$ is the Hamiltonian of unit 14.

The proof, written without the compression. Set $E = F - y'F_{y'}$, where $F$ and its derivatives are evaluated at $(x, y(x), y'(x))$. Then

$$\frac{\mathrm dE}{\mathrm dx} = \underbrace{F_x + F_yy' + F_{y'}y''}_{\mathrm dF/\mathrm dx} \;-\; \underbrace{y''F_{y'} - y'\frac{\mathrm d}{\mathrm dx}F_{y'}}_{\mathrm d(y'F_{y'})/\mathrm dx} \;=\; F_x + y'\Big(F_y - \frac{\mathrm d}{\mathrm dx}F_{y'}\Big).$$

The bracket is zero on an extremal, leaving $\mathrm dE/\mathrm dx = F_x$, hence $E$ is constant when $F_x = 0$. The combination $H = y'F_{y'} - F = -E$ is the one that becomes the Hamiltonian, and it obeys $\mathrm dH/\mathrm dx = -F_x$; the two are the same statement with opposite signs, so use one consistently.

The converse fails, and it matters in section 3. The computation shows $\mathrm dE/\mathrm dx = F_x + y'\,\mathcal E$, where $\mathcal E = F_y - (F_{y'})'$ is the Euler–Lagrange expression. So a constant $E$ forces $\mathcal E = 0$ only where $y' \ne 0$. A curve with $y' \equiv 0$ on an interval satisfies the Beltrami identity whatever $\mathcal E$ is there; such solutions must be checked against the full equation before being believed.

Degenerate case. The identity is vacuous when $F_{y'y'} = 0$ identically, because then $F - y'F_{y'}$ is constant for every curve, extremal or not, and the "first integral" carries no information. Check it on $F = y'$: $F - y'F_{y'} = 0$ always. This is the same degeneracy flagged two slides ago.

Attribution. The identity carries Eugenio Beltrami's name; the first integral itself was already used by Euler. It is also, read backwards, the reason unit 14 defines the Hamiltonian by a Legendre transform: $H = y'F_{y'} - F$ is exactly the Legendre transform of $F$ in the variable $y'$, and the identity says the transform is conserved when the independent variable is absent.

section 3

Solving it: the cycloid

Beltrami turns the bead's second-order equation into a first-order one, and a substitution integrates it.

11 · the cycloid

Beltrami on the bead

In the descent-time functional, $F = \sqrt{1+p^2}\,/\sqrt{2gy}$ with $p = y'$, and $x$ does not appear. So Beltrami applies.

Step 1. $F_{y'} = \dfrac{y'}{\sqrt{1+y'^2}\sqrt{2gy}}$.

Step 2. $F - y'F_{y'} = \dfrac{(1+y'^2) - y'^2}{\sqrt{1+y'^2}\sqrt{2gy}} = \dfrac{1}{\sqrt{1+y'^2}\sqrt{2gy}}$.

Step 3. Call the constant $1/\sqrt{4ga}$, with $a > 0$ a length to be found. Squaring and inverting:

the first integral $$y\,\big(1+y'^2\big) = 2a$$

dimensions

$F - y'F_{y'}$ is in $\mathrm{s\,m^{-1}}$, and so is $1/\sqrt{4ga}$ exactly when $a$ is a length; then $y(1+y'^2) = 2a$ is metres on both sides ✓.

a spurious solution

$y \equiv 2a$ satisfies the first integral but is no extremal: there $F_y \ne 0 = (F_{y'})'$.

The Beltrami proof showed $\frac{\mathrm d}{\mathrm dx}\big(F - y'F_{y'}\big) = F_x + y'\big(F_y - (F_{y'})'\big)$. Where $y' = 0$ the Euler–Lagrange bracket is multiplied by zero, so a constant Beltrami quantity says nothing about it there. Such solutions must be checked against the full equation.

11 · the cycloid

The first integral is Snell's law in layers

read it physically

Let $\phi$ be the angle between the wire and the vertical. The tangent is along $(1, y')$, so $\sin\phi = 1/\sqrt{1+y'^2}$, and $v = \sqrt{2gy}$. The first integral says

$$\frac{\sin\phi}{v} = \frac{1}{\sqrt{4ga}} = \text{const},$$

which is Snell's law for horizontal layers of speed $v(y)$. Where $\phi = 90^\circ$ the wire is horizontal and $v$ is largest: $2a$ is the depth of the lowest point.

The running example's cycloid, $a = 0.5172$ m, across layers of increasing speed. Like light entering faster glass, it bends away from the vertical as it descends — $\sin\phi \propto \sqrt y$ — until it runs flat at depth $2a$, then climbs to $B$.

Reading the figure. At each marked point the short segment is the tangent and the dashed line the vertical; $\phi$ is the angle between them. From the first integral, $\sin\phi = v/\sqrt{4ga} = \sqrt{y/(2a)}$, so $\phi$ is fixed by the depth alone: $\phi = 90^\circ$ exactly at $y = 2a$. On the climb back to $B$ the same depths give the same angles, mirrored — the curve is symmetric about its lowest point.

Johann Bernoulli's route, now justified. He argued: in a thin horizontal layer the speed is constant, so the quickest path crosses it as light would, and Snell at every interface gives $\sin\phi/v$ constant. The derivation on the previous slide is the same statement with no layers and no limit, and it shows that the answer does not depend on believing that light-like refraction is the right model for a bead. Unit 15 explains why the analogy was bound to work: mechanics and optics are the same variational problem.

11 · the cycloid · example

Worked: integrating the first integral

from $y(1+y'^2) = 2a$ to a curve

Step 1. Solve for the slope: $y'^2 = (2a - y)/y$. On the descending part $y' > 0$, and unit 00's separation of variables gives $\ \mathrm dx = \sqrt{y/(2a-y)}\;\mathrm dy$.

Step 2. The square root wants a half-angle. Put $y = 2a\sin^2(\theta/2) = a(1-\cos\theta)$, $\theta\in(0,\pi)$. Then $2a - y = 2a\cos^2(\theta/2)$, so $\sqrt{y/(2a-y)} = \tan(\theta/2)$, and $\mathrm dy = a\sin\theta\,\mathrm d\theta = 2a\sin(\theta/2)\cos(\theta/2)\,\mathrm d\theta$.

Step 3. Multiply: $\ \mathrm dx = \tan(\theta/2)\cdot 2a\sin(\theta/2)\cos(\theta/2)\,\mathrm d\theta = 2a\sin^2(\theta/2)\,\mathrm d\theta = a(1-\cos\theta)\,\mathrm d\theta$.

Step 4. Integrate, $x = a(\theta - \sin\theta) + x_0$; the curve starts at $A$, where $\theta = 0$, so $x_0 = 0$.

$$x = a(\theta - \sin\theta), \qquad y = a(1 - \cos\theta).$$

checks

Back-substitute. $y' = \dfrac{\sin\theta}{1-\cos\theta} = \cot\tfrac\theta2$, so $y(1+y'^2) = \dfrac{2a\sin^2(\theta/2)}{\sin^2(\theta/2)} = 2a$ ✓.

At $A$. $x \approx a\theta^3/6$, $y \approx a\theta^2/2$: the wire leaves vertically, as the singular-start slide predicted.

Past the bottom. The substitution was made for $\theta\in(0,\pi)$, where $y' > 0$. The parametric curve continues smoothly through $\theta = \pi$, where $y' = \cot(\pi/2) = 0$, and on $(\pi, 2\pi)$ it has $y' = \cot(\theta/2) < 0$ and still satisfies $y(1+y'^2) = 2a$ — that is Step 1 with the other sign of the square root. Is it still an extremal there? On each open piece where $y' \ne 0$ the Beltrami constant forces the Euler–Lagrange expression $\mathcal E$ to vanish (the converse remark under the Beltrami slide), and $\mathcal E$ is continuous in $x$, so it also vanishes at the single point $\theta = \pi$. Hence the whole arch is an extremal. The running example needs this, because its endpoint sits at $\theta_B = 3.508 > \pi$.

The expansion at $A$. $\theta - \sin\theta = \theta^3/6 - \theta^5/120 + \cdots$ and $1 - \cos\theta = \theta^2/2 - \cdots$, so eliminating $\theta$ gives $y \approx (9a/2)^{1/3}x^{2/3}$. The slope $y' \approx \tfrac23(9a/2)^{1/3}x^{-1/3} \to \infty$: vertical departure, the fastest possible initial acceleration, bought at the price of a longer path.

11 · the cycloid

Why the answer is called a cycloid

the answer has a name

The curve $x = a(\theta - \sin\theta)$, $y = a(1 - \cos\theta)$ is a cycloid: the path of a point on the rim of a circle of radius $a$ rolling along the underside of the line $y = 0$, $\theta$ being the angle it has turned.

After turning through $\theta$ the circle has rolled $a\theta$, so its centre is at $(a\theta, a)$ and the rim point at $(a\theta, a) - a(\sin\theta, \cos\theta)$ — the pair just derived. The running example uses the solid part, $\theta$ from $0$ to $\theta_B$, past the bottom.

The rolling circle. A circle of radius $a$ rolling under the line $y = 0$ has its centre at $(a\theta, a)$ after turning through $\theta$ (no slipping: the arc rolled equals the distance travelled). A rim point that started at the origin — the point of contact at $\theta = 0$ — sits at $(a\theta, a) - a(\sin\theta, \cos\theta)$, which is the pair above. Nothing in the proof uses this picture; it is only why the curve has a name, and why Pascal, Galileo and Huygens had all studied it before 1696.

What to see in it. The rim point moves fastest when it is at the bottom of the circle, furthest from the contact point, and is momentarily at rest at the top, where it touches the line: so the cycloid meets $y = 0$ in a cusp, vertically, as the checks on the previous slide found at $A$. The figure is drawn with $a = 1$; every cycloid is this one scaled.

11 · the cycloid

The constants, and the race

Fixing $\theta_B$. At $B$, dividing $x$ by $y$ removes $a$:

$$r(\theta_B) := \frac{\theta_B - \sin\theta_B}{1 - \cos\theta_B} = \frac{X}{Y} = 2 .$$

$r$ increases strictly from $0$ to $\infty$ on $(0, 2\pi)$ (notes), so there is exactly one root: $\theta_B = 3.50837$, i.e. $201.01^\circ$. Then $a = Y/(1-\cos\theta_B) = 0.51720$ m.

The time. Along the arch, $\mathrm ds = 2a\sin(\theta/2)\,\mathrm d\theta$ and $v = \sqrt{2gy} = 2\sqrt{ga}\sin(\theta/2)$:

$$\mathrm dt = \sqrt{a/g}\;\mathrm d\theta, \qquad T = \theta_B\sqrt{a/g} = 0.80556\ \mathrm s .$$

The straight line is $25.3\%$ slower, Galileo's arc $3.07\%$. The cycloid dips $3.44$ cm below $B$ and climbs back.

Full proof that $r$ is strictly increasing

With $r(\theta) = (\theta - \sin\theta)/(1-\cos\theta)$ on $(0, 2\pi)$, the quotient rule gives

$$r'(\theta) = \frac{(1-\cos\theta)^2 - (\theta - \sin\theta)\sin\theta}{(1-\cos\theta)^2} = \frac{2 - 2\cos\theta - \theta\sin\theta}{(1-\cos\theta)^2} = \frac{2 - \theta\cot(\theta/2)}{1 - \cos\theta},$$

using $\sin\theta/(1-\cos\theta) = \cot(\theta/2)$. Write $h(\theta) = 2 - \theta\cot(\theta/2)$. For $\theta\in(0,\pi)$, put $u = \theta/2 \in (0, \pi/2)$: then $h = 2(1 - u/\tan u) > 0$ because $\tan u > u$ there. For $\theta\in[\pi, 2\pi)$, $\cot(\theta/2) \le 0$, so $h \ge 2$. Hence $r' > 0$ throughout. The limits: $r(\theta) \approx \theta/3 \to 0$ as $\theta\to0$, and $r \to \infty$ as $\theta \to 2\pi$ because the denominator vanishes and the numerator tends to $2\pi$. So $r$ is a bijection $(0,2\pi)\to(0,\infty)$ and every $B$ with $X, Y > 0$ is reached by exactly one first arch. Keep $h$ in mind: section 5 meets it again as a Jacobi field, and that is not a coincidence.

The competitors. Straight line: the bead has constant acceleration $g\,Y/L$ along the incline of length $L = \sqrt{X^2+Y^2}$, so $L = \tfrac12 (gY/L)T^2$ and $T = \sqrt{2(X^2+Y^2)/(gY)} = \sqrt{10/9.81} = 1.00964$ s. Circular arc: the circle centred on the $x$-axis through $A$ and $B$ leaves $A$ vertically; its centre $(R,0)$ satisfies $(2-R)^2 + 1 = R^2$, so $R = 1.25$ m. Parametrise by the angle $\psi$ from the centre, $x = R - R\cos\psi$, $y = R\sin\psi$, with $\mathrm ds = R\,\mathrm d\psi$ and $v = \sqrt{2gR\sin\psi}$; then $T = \int_0^{\psi_B}\sqrt{R/(2g\sin\psi)}\,\mathrm d\psi$ with $\cos\psi_B = (R-2)/R$, $\psi_B = 126.87^\circ$. The integrand has an integrable $\psi^{-1/2}$ singularity at $0$ (unit 00), and quadrature gives $0.83031$ s.

The cycloid's time, line by line. $\mathrm dx = a(1-\cos\theta)\,\mathrm d\theta$ and $\mathrm dy = a\sin\theta\,\mathrm d\theta$ give $\mathrm ds = a\sqrt{2 - 2\cos\theta}\,\mathrm d\theta = 2a\sin(\theta/2)\,\mathrm d\theta$ for $\theta\in(0,2\pi)$. And $2gy = 2ga(1-\cos\theta) = 4ga\sin^2(\theta/2)$. The ratio is $\sqrt{a/g}$, constant: the parameter $\theta$ is time in units of $\sqrt{a/g} = 0.22961$ s. So $T = 3.50837 \times 0.22961 = 0.80556$ s. The lowest point, $2a = 1.03440$ m deep, is $3.44$ cm below $B$: the quickest wire overshoots and climbs, which no intuition about short paths would suggest.

11 · the cycloid

Checking the answer from both ends

dimensions

$a = Y/(1 - \cos\theta_B)$ is a length; $\sqrt{a/g}$ is a time; $\theta_B$ is a pure number. So $T$ is seconds ✓. And, as promised by the mass-free functional, $m$ appears nowhere.

limit 1 · $B$ straight below $A$

As $X/Y\to0$: $r(\theta)\approx\theta/3$ gives $\theta_B \approx 3X/Y \to 0$, and $a \approx 2Y/\theta_B^2$. So

$$T = \theta_B\sqrt{a/g} \;\to\; \sqrt{2Y/g},$$

the free-fall time. The cycloid degenerates to the vertical segment ✓.

limit 2 · $B$ level with $A$

As $Y\to0$ with $X$ fixed: $\theta_B\to2\pi$ and $a \to X/(2\pi)$, so $T \to \sqrt{2\pi X/g}$, finite. For $X = 2$ m that is $1.13180$ s: the bead runs a full arch and arrives with zero speed, as energy conservation demands ✓.

a bonus: the tautochrone

Released from rest anywhere on the arch, the bead reaches the bottom in the same time $\pi\sqrt{a/g}$ — for the running example's cycloid, $0.72135$ s. Huygens, 1673; the proof is three lines, in the notes.

Limit 1 in detail. For small $\theta$, $\theta - \sin\theta = \theta^3/6 + O(\theta^5)$ and $1 - \cos\theta = \theta^2/2 + O(\theta^4)$, so $r(\theta) = \theta/3 + O(\theta^3)$ and $\theta_B = 3X/Y + O((X/Y)^3)$. Then $a = Y/(1-\cos\theta_B) = 2Y/\theta_B^2\,(1 + O(\theta_B^2))$ and $T = \theta_B\sqrt{a/g} = \sqrt{2Y/g}\,(1 + O(\theta_B^2))$. So the free-fall time is recovered, with a correction of relative size $(X/Y)^2$.

Limit 2 in detail. As $\theta\to2\pi$, $1 - \cos\theta\to0$ while $\theta - \sin\theta \to 2\pi$, so $X/Y = r(\theta_B)\to\infty$ is consistent with $Y\to 0$, and $X = a(\theta_B - \sin\theta_B) \to 2\pi a$. Then $T\to 2\pi\sqrt{X/(2\pi g)} = \sqrt{2\pi X/g} = \sqrt{4\pi/9.81} = 1.13180$ s.

Full proof of the tautochrone property

Measure along the arch from its lowest point, $\theta = \pi$. Put $\sigma = \theta - \pi$. Arc length from the bottom is $s = \int_\pi^\theta 2a\sin(t/2)\,\mathrm dt = -4a\cos(\theta/2) = 4a\sin(\sigma/2)$, and the height above the bottom is $h = 2a - y = a(1+\cos\theta) = 2a\cos^2(\theta/2) = s^2/(8a)$. So the potential energy per unit mass is $g s^2/(8a)$, and energy conservation along the wire from rest at $s_0$ reads $\tfrac12\dot s^2 + g s^2/(8a) = g s_0^2/(8a)$, i.e. $\dot s^2 = \omega^2(s_0^2 - s^2)$ with $\omega = \sqrt{g/(4a)}$: simple harmonic motion in $s$, whatever the amplitude. Unit 00's quadrature gives the time to the bottom, $\int_0^{s_0}\mathrm ds/\big(\omega\sqrt{s_0^2 - s^2}\big) = \big[\arcsin(s/s_0)\big]_0^{s_0}/\omega = \pi/(2\omega)$, independent of $s_0$: that is $\tfrac{\pi}{2}\sqrt{4a/g} = \pi\sqrt{a/g}$ — which for $a = 0.51720$ m is $0.72135$ s. Huygens built pendulum clocks with cycloidal cheeks to exploit this, since a circular pendulum's period drifts with amplitude (unit 00's elliptic integral).

section 4

Constraints and free ends

An end left loose, a length held fixed, a curve kept on a surface: each changes which variations are allowed, and so what the first variation must kill.

11 · constraints

A free end: the natural boundary condition

theorem · natural boundary condition

Fix $y(a) = \alpha$ and leave $y(b)$ free. If $y\in C^2$ is an extremal among such curves, it satisfies the Euler–Lagrange equation on $[a,b]$ and

$$F_{y'}\big(b, y(b), y'(b)\big) = 0 .$$

proof

Now $\eta$ need only vanish at $a$. Integrating by parts, $\delta J = \int_a^b\big(F_y - (F_{y'})'\big)\eta\,\mathrm dx + F_{y'}\big|_b\,\eta(b)$. Variations with $\eta(b) = 0$ are still allowed, so the fundamental lemma gives Euler–Lagrange and the integral vanishes. Then $\eta(x) = (x-a)/(b-a)$ leaves $F_{y'}|_b = 0$. ∎

Two boundary conditions, as a second-order equation needs: one imposed at $a$, one produced at $b$ by the minimisation itself.

Why "natural". Nobody imposes $F_{y'} = 0$ at $b$. The minimisation produces it, as the price of leaving the end free: a curve that ended with $F_{y'} \ne 0$ could be improved by moving its end. The two boundary conditions a second-order equation needs are therefore always there — one imposed at $a$, one natural at $b$.

The integration by parts, carefully. With $y\in C^2$ and $F\in C^2$, the function $x\mapsto F_{y'}(x,y(x),y'(x))$ is $C^1$, so $\int_a^b F_{y'}\eta'\,\mathrm dx = [F_{y'}\eta]_a^b - \int_a^b (F_{y'})'\eta\,\mathrm dx$, and $\eta(a) = 0$ removes the lower boundary term. The test function $(x-a)/(b-a)$ vanishes at $a$ and equals $1$ at $b$; any $C^1$ function with those values would do.

The physics of the condition is a force balance, and problem P5 makes it concrete: a string whose end is on a ring sliding freely on a vertical rod must meet the rod at right angles, because the ring can exert no vertical force.

11 · constraints

A free end, on the bead

on the bead: reach the line $x = X$ soonest

$F_{y'} = y'/\sqrt{2gy(1+y'^2)} = 0$ at $x = X$ forces $y'(X) = 0$: the wire meets the line horizontally, at the bottom of the arch, $\theta = \pi$. So $X = a\pi$ and

$$T = \pi\sqrt{a/g} = \sqrt{\pi X/g} .$$

For $X = 2$ m: $a = 0.63662$ m, landing depth $2a = 1.27324$ m, $T = 0.80030$ s — $0.0053$ s better than aiming at $B$, as a larger admissible set must allow.

Checks. $\sqrt{X/g}$ is seconds ✓; $T$ grows like $\sqrt X$, the scaling of unit 00's mechanical similarity ✓.

Let the landing depth go free and the best cycloid gets bigger: it reaches the line at its lowest, fastest point instead of spending the end of the trip climbing back up to $B$.

11 · constraints

An integral constraint: a multiplier in function space

theorem · the isoperimetric problem

Let $y$ be a local minimiser of $J[y] = \int_a^b F\,\mathrm dx$ among curves in $\mathcal A$ with $K[y] = \int_a^b G(x,y,y')\,\mathrm dx = \ell$, and suppose $y$ is not an extremal of $K$. Then there is a constant $\lambda$ with

$$\delta J[y;\eta] = \lambda\,\delta K[y;\eta]\ \ \text{for all }\eta, \quad\text{so}\quad \frac{\mathrm d}{\mathrm dx}\big(F-\lambda G\big)_{y'} = \big(F - \lambda G\big)_y .$$

proof · the idea

Pick $\eta_2$ with $\delta K[y;\eta_2]\ne0$. For any $\eta_1$, the two-variable function $(\varepsilon_1,\varepsilon_2)\mapsto J[y + \varepsilon_1\eta_1 + \varepsilon_2\eta_2]$ has a local minimum at $0$ on the curve where $K = \ell$. Unit 04's Lagrange multiplier theorem applies in $\mathbb R^2$. Full proof in the notes.

what $\lambda$ is

If the constrained minimum $J^*(\ell)$ depends smoothly on $\ell$, then $\lambda = \mathrm dJ^*/\mathrm d\ell$: the price of one more unit of constraint, exactly as in unit 04. Dimensions: $[\lambda] = [J]/[K]$ ✓.

Full proof

Fix $\eta_2 \in C^1_0$ with $\delta K[y;\eta_2] \ne 0$, which exists because $y$ is not an extremal of $K$. Let $\eta_1\in C^1_0$ be arbitrary and define on a neighbourhood of $0\in\mathbb R^2$

$$\Phi(\varepsilon_1,\varepsilon_2) = J[y + \varepsilon_1\eta_1 + \varepsilon_2\eta_2],\qquad \Psi(\varepsilon_1,\varepsilon_2) = K[y + \varepsilon_1\eta_1 + \varepsilon_2\eta_2].$$

The argument of the first-variation slide, applied to each partial derivative, shows that $\Phi$ and $\Psi$ are $C^1$, with $\partial_{\varepsilon_i}\Phi(0) = \delta J[y;\eta_i]$ and $\partial_{\varepsilon_i}\Psi(0) = \delta K[y;\eta_i]$. Every point of $\{\Psi = \ell\}$ near $0$ gives an admissible curve satisfying the constraint, so $\Phi$ restricted to that set has a local minimum at $0$. Since $\nabla\Psi(0) \ne 0$ (its second component is non-zero), unit 04's theorem gives a $\lambda$ with $\nabla\Phi(0) = \lambda\nabla\Psi(0)$. The second component reads $\delta J[y;\eta_2] = \lambda\,\delta K[y;\eta_2]$, so $\lambda = \delta J[y;\eta_2]/\delta K[y;\eta_2]$ — which does not involve $\eta_1$. The first component then reads $\delta J[y;\eta_1] = \lambda\,\delta K[y;\eta_1]$ for every $\eta_1$ with the same $\lambda$. That says $F - \lambda G$ has vanishing first variation, and section 2 turns that into its Euler–Lagrange equation.

Full proof that $\lambda = \mathrm dJ^*/\mathrm d\ell$

Suppose the constrained minimiser $y_\ell$ depends on $\ell$ so that $\partial_\ell y_\ell$ exists, is $C^1$ in $x$, and may be differentiated under the integral. Since every $y_\ell$ has the same endpoint values, $\eta = \partial_\ell y_\ell$ vanishes at $a$ and $b$ and is an admissible variation. The chain rule gives $\frac{\mathrm d}{\mathrm d\ell}J[y_\ell] = \delta J[y_\ell;\eta] = \lambda\,\delta K[y_\ell;\eta] = \lambda\frac{\mathrm d}{\mathrm d\ell}K[y_\ell] = \lambda$, because $K[y_\ell] = \ell$. The smooth-dependence hypothesis is real and is assumed, not proved.

The exceptional case. If $y$ is an extremal of $K$, the theorem can fail: the constraint set may be too thin near $y$ for its tangent directions to be all of $\{\eta : \delta K[y;\eta] = 0\}$. The general statement puts a multiplier on $J$ too, $\lambda_0\,\delta J = \lambda\,\delta K$ with $(\lambda_0,\lambda)\ne0$ — Gelfand and Fomin, §12, cited.

The name. The oldest problem of this kind is Dido's — enclose the most area with a given length of boundary — and it is problem P6.

11 · constraints · example

Worked: the hanging chain

least potential energy at fixed length

A chain of length $\ell$ and mass per length $\rho$ hangs from $(\pm d, 0)$; here $y$ is measured upwards. Minimise $U = \rho g\!\int y\sqrt{1+y'^2}\,\mathrm dx$ with $\int\sqrt{1+y'^2}\,\mathrm dx = \ell$.

Step 1. By the theorem, extremise $\int(\rho g y - \mu)\sqrt{1+y'^2}\,\mathrm dx$ for some constant $\mu$. No explicit $x$, so Beltrami: $\ \dfrac{\rho g y - \mu}{\sqrt{1+y'^2}} = C$.

Step 2. Shift and scale: $u = y - \mu/(\rho g)$ and $c = C/(\rho g)$ give $u^2 = c^2(1+u'^2)$, so $u' = \pm\sqrt{u^2/c^2 - 1}$.

Step 3. Substitute $u = c\cosh s$: then $\sqrt{u^2/c^2-1} = \sinh s$ and $\mathrm du = c\sinh s\,\mathrm ds$, so $c\,\mathrm ds = \mathrm dx$ and $u = c\cosh\big((x - x_0)/c\big)$: a catenary.

Step 4. Symmetry of the supports gives $x_0 = 0$. The length fixes $c$, and $y(\pm d) = 0$ fixes $\mu$:

$$2c\sinh(d/c) = \ell, \qquad \mu = -\rho g\,c\cosh(d/c).$$

numbers and checks

$d = 1$ m, $\ell = 3$ m, $\rho = 2\ \mathrm{kg\,m^{-1}}$: $c = 0.61647$ m, sag $c(\cosh(d/c) - 1) = 1.00527$ m.

Dimensions. $c$ is a length; $\rho g c$ is a force ✓.

Limit. $\ell\downarrow 2d$ forces $c\to\infty$ and the sag to $0$: a taut straight chain ✓.

Weight. At each support the vertical pull is $\rho g c\sinh(d/c) = \rho g\ell/2 = 29.43$ N, half the weight ✓.

Beltrami for the combined integrand. With $\tilde F = (\rho g y - \mu)\sqrt{1+p^2}$, $\tilde F_p = (\rho g y - \mu)\,p/\sqrt{1+p^2}$, and $\tilde F - p\tilde F_p = (\rho g y - \mu)\big[(1+p^2) - p^2\big]/\sqrt{1+p^2} = (\rho g y-\mu)/\sqrt{1+p^2}$. Is the chain an extremal of the length constraint $K$ alone, which would void the theorem? Extremals of length are straight lines, and a chain with $\ell > 2d$ is not one, so the hypothesis holds.

Step 3 in detail. On a piece where $u' > 0$, separation gives $\mathrm dx = c\,\mathrm du/\sqrt{u^2 - c^2}$; the substitution $u = c\cosh s$ (with $s > 0$) makes the right side $c\,\mathrm ds$, so $s = (x-x_0)/c$. The other sign gives $s < 0$, and the single formula $u = c\cosh((x-x_0)/c)$ covers both, joined smoothly at $x = x_0$ where $u' = 0$. The constant $c$ must be positive: $u = \sqrt{1+u'^2}\,c$ has the sign of $c$, and the tension argument on the next slide shows $u > 0$. Why $x_0 = 0$: $y(d) = y(-d)$ means $\cosh((d-x_0)/c) = \cosh((d+x_0)/c)$; $\cosh$ is even and strictly increasing on $[0,\infty)$, so $|d - x_0| = |d + x_0|$, whence $x_0 = 0$.

Solving for $c$. $2c\sinh(1/c) = 3$ has exactly one positive root, because $c\mapsto 2c\sinh(d/c)$ decreases strictly from $\infty$ to $2d$ on $(0,\infty)$; numerically $c = 0.61647$ m. Then $\cosh(d/c) = 2.6307$ and the sag is $1.00527$ m.

History. Galileo (1638) said the chain hangs in a parabola; Huygens showed at 17 that it does not; Leibniz, Huygens and Johann Bernoulli found the curve in 1691, answering Jacob Bernoulli's challenge. Leibniz's solution is effectively this one; the multiplier had to wait for Euler and Lagrange.

11 · constraints

The multiplier is the tension

claim

The tension in the chain at height $y$ is $\ T = \rho g\,y - \mu$. At the supports, where $y = 0$, $\ T = -\mu$.

proof

No horizontal force acts on the chain except at its ends, so the horizontal component $H$ of the tension is the same everywhere; at the bottom it is all of the tension. A piece from the bottom to $x$ is held by its weight $\rho g s(x)$, $s = c\sinh(x/c)$ the arc length, so the vertical component there is $\rho g c\sinh(x/c)$. Dividing, $\tan(\text{angle}) = y' = \sinh(x/c)$ forces $H = \rho g c$, and $T = H\sqrt{1+y'^2} = \rho g c\cosh(x/c) = \rho g u = \rho g y - \mu$. ∎

The free body in the proof: the piece from the bottom to $x$. Horizontally only $H$ and the horizontal part of $T$ act, so they balance; vertically the weight of the arc $s(x)$ balances the vertical part of $T$.

The force balance, more carefully. The chain is flexible, so the force one part exerts on the next is a tension along the tangent. Cut at the lowest point $x=0$ and at $x>0$. The piece in between is in equilibrium under: the pull $H$ at the bottom, horizontal and pointing towards negative $x$; its weight $\rho g s(x)$ downward; and the tension $T(x)$ along the tangent at $x$, with components $(T\cos\alpha, T\sin\alpha)$ and $\tan\alpha = y'(x)$. Horizontal balance gives $T\cos\alpha = H$; vertical gives $T\sin\alpha = \rho g s(x)$, with $s(x) = \int_0^x\cosh(t/c)\,\mathrm dt = c\sinh(x/c)$. Dividing: $\sinh(x/c) = y' = \rho g c\sinh(x/c)/H$, so $H = \rho g c$. Then $T = H/\cos\alpha = H\sqrt{1 + y'^2} = \rho g c\cosh(x/c)$.

The general moral, which unit 12 uses on every constrained system: a Lagrange multiplier for a constraint is the force that enforces it. Here the constraint is inextensibility, and the force is tension.

11 · constraints

The tension, checked, and Galileo's parabola priced

the numbers agree

$T_{\text{support}} = \rho g c\cosh(d/c) = 31.8185$ N. Differentiating the minimum energy $U^*(\ell)$ numerically at $\ell = 3$ m gives $\mathrm dU^*/\mathrm d\ell = -31.8185$ N $= \mu$ ✓.

Why. Draw $\delta\ell$ of chain out through a support against tension $T$: the work $T\,\delta\ell$ goes into the hanging part, so $\mathrm dU^*/\mathrm d(-\ell) = T$.

Galileo's parabola, priced

The parabola through the supports with the same length, $y = 1.02768\,(x^2 - 1)$, has $U = -35.5705$ J; the catenary has $U^* = -35.6326$ J. Lower by $0.0620$ J — small, and of the right sign.

Same supports, same $3$ m of chain. The parabola hangs $2.24$ cm lower in the middle but up to $4.0$ cm higher near the ends, where it has had to take the length from; the catenary's mass sits lower on balance.

Why the sign. $\mu$ is negative, the tension positive. In the course's sign convention the extremised integrand was $\rho g y - \mu$ per unit length: the multiplier $\mu$ is the price $\mathrm dU^*/\mathrm d\ell$ of length, and extra length lowers the energy. The isoperimetric slide's sensitivity result says $\mu = \mathrm dU^*/\mathrm d\ell$, and the energy argument says that rate is $-T_{\text{support}}$; the force balance says the same thing independently, which is the point of computing it twice.

How the numbers were made. $U^*(\ell) = \rho g\int_{-d}^{d}(y_0 + c\cosh(x/c)) \cosh(x/c)\,\mathrm dx$ with $y_0 = -c\cosh(d/c)$ and $c$ solving $2c\sinh(d/c) = \ell$, differentiated numerically at $\ell = 3$ m; it agrees with $-\rho g c\cosh(d/c)$ to every digit shown. The parabola $y = k(x^2 - 1)$ has length $\int_{-1}^{1}\sqrt{1 + 4k^2x^2}\,\mathrm dx = 3$ at $k = 1.02768$, and its energy $\rho g\int_{-1}^1 k(x^2-1)\sqrt{1+4k^2x^2}\,\mathrm dx = -35.5705$ J. Both energies are negative because they are measured from the supports. The lower panel of the figure is $1.02768(x^2-1) - y_{\text{cat}}(x)$ in centimetres, with $y_{\text{cat}} = c\cosh(x/c) - c\cosh(1/c)$: $-2.24$ cm at $x = 0$ and $+4.03$ cm at its largest, near $x = \pm0.8$ m.

11 · constraints

Several unknowns, and higher derivatives

several dependent variables

For $J[\mathbf q] = \int_{t_0}^{t_1}F(t, \mathbf q, \mathbf q')\,\mathrm dt$ with $\mathbf q = (q_1,\dots,q_n)$ and both ends fixed, an extremal satisfies $n$ equations

$$\frac{\mathrm d}{\mathrm dt}F_{q_i'} = F_{q_i},\qquad i = 1,\dots,n.$$

Proof. Vary one component at a time, $\boldsymbol\eta = \eta\,\mathbf e_i$, and apply the one-variable theorem. ∎

This is the form unit 12 uses for $n$ degrees of freedom. It also frees curves from being graphs: write $(x(t), y(t))$ and the vertical tangent is no longer special.

higher derivatives · the Euler–Poisson equation

For $J = \int_a^b F(x, y, y', y'')\,\mathrm dx$ with $y$ and $y'$ fixed at both ends, a $C^4$ extremal satisfies

$$F_y - \frac{\mathrm d}{\mathrm dx}F_{y'} + \frac{\mathrm d^2}{\mathrm dx^2}F_{y''} = 0 .$$

Proof. Now $\eta$ and $\eta'$ vanish at $a$, $b$. Integrate $F_{y'}\eta'$ by parts once and $F_{y''}\eta''$ twice; every boundary term dies. The fundamental lemma finishes. ∎

Each term has dimension $[F]/[y]$ ✓. With $F = \tfrac12 EI\,y''^2 + \rho g y$ it is the bending equation of a beam: problem P10.

Several variables, in detail. With $\boldsymbol\eta = \eta\,\mathbf e_i$, the first variation is $\int(F_{q_i}\eta + F_{q_i'}\eta')\,\mathrm dt$, word for word the one-variable expression with the other components frozen along the extremal. Section 2's argument gives the $i$-th equation. Conversely, a general $\boldsymbol\eta$ is a sum of such, and the first variation is linear, so the $n$ equations are also sufficient for stationarity.

A warning about parametric curves. For arc length $F = \sqrt{x'^2 + y'^2}$ the two equations are not independent: $F$ is homogeneous of degree one in $(x', y')$, so the length is unchanged by reparametrisation, and any reparametrisation of an extremal is again one. The matrix $F_{q'q'}$ is singular in the direction of motion, and the two equations carry the content of one. Unit 27 meets exactly this with the relativistic free particle; the usual cure — used on the next slide — is to extremise $\tfrac12\int\lvert\mathbf q'\rvert^2$ instead, whose extremals are the extremals of length traversed at constant speed.

Full proof of the Euler–Poisson equation

Assume $F\in C^3$ and $y\in C^4$, so that $F_{y'}$ is $C^1$ and $F_{y''}$ is $C^2$ as functions of $x$ along $y$. Admissible variations are $\eta \in C^2$ with $\eta(a) = \eta'(a) = \eta(b) = \eta'(b) = 0$. Then $\delta J = \int_a^b(F_y\eta + F_{y'}\eta' + F_{y''}\eta'')\,\mathrm dx$. Integrating by parts, $\int F_{y'}\eta' = [F_{y'}\eta] - \int (F_{y'})'\eta$ and $\int F_{y''}\eta'' = [F_{y''}\eta'] - [(F_{y''})'\eta] + \int (F_{y''})''\eta$, and every bracket vanishes. So $\int_a^b\big(F_y - (F_{y'})' + (F_{y''})''\big)\eta\,\mathrm dx = 0$ for all such $\eta$. The bump $\eta_{c,d}$ of section 2 has a triple root at each end of its support, so it and $\eta'$ vanish at $a$ and $b$: it is admissible, and the fundamental lemma's proof goes through unchanged. Poisson wrote the equation down in 1833; for $F$ depending on $y^{(k)}$ the pattern continues with alternating signs.

11 · constraints

A pointwise constraint: curves on a surface

theorem · a holonomic constraint

Let $\mathbf q:[t_0,t_1]\to\mathbb R^3$ be a $C^2$ extremal of $\int F(t,\mathbf q,\mathbf q')\,\mathrm dt$ among curves on the surface $G(\mathbf q) = 0$ with fixed ends, and $\nabla G \ne 0$ along it. Then there is a continuous function $\lambda(t)$ with

$$\frac{\mathrm d}{\mathrm dt}F_{\mathbf q'} - F_{\mathbf q} = \lambda(t)\,\nabla G(\mathbf q(t)) .$$

implicit function theorem to solve $G = 0$ locally, then the chain rule — full proof in the notes (N)

on the sphere $\lvert\mathbf q\rvert = R$

Take $F = \tfrac12\lvert\mathbf q'\rvert^2$ and $G = \lvert\mathbf q\rvert^2 - R^2$. Then $\mathbf q'' = 2\lambda\,\mathbf q$, so $(\mathbf q\times\mathbf q')' = \mathbf q\times\mathbf q'' = \mathbf 0$ and $\mathbf q$ stays in the fixed plane through $0$ normal to $\mathbf q\times\mathbf q'$: a great circle.

A bead on a frictionless sphere, no gravity. Its acceleration $\mathbf q''$ is all along the normal $\nabla G$, with no sideways part, so the plane of $\mathbf q$ and $\mathbf q'$ never turns.

Full proof

Work near a point of the curve where $G_z \ne 0$ (some partial is non-zero; relabel). By unit 04's implicit function theorem the surface is locally $z = h(x,y)$ with $h_x = -G_x/G_z$ and $h_y = -G_y/G_z$. Curves on the surface are then given by free $(x(t),y(t))$, and $J$ becomes $\tilde J[x,y] = \int F\big(t, x, y, h, x', y', h_xx' + h_yy'\big)\,\mathrm dt$, an unconstrained problem in two unknowns, so the previous slide gives its two Euler–Lagrange equations. Write $E_i = F_{q_i} - \frac{\mathrm d}{\mathrm dt}F_{q_i'}$ for the Euler expressions of the original $F$. The chain rule gives $\tilde F_x = F_x + F_zh_x + F_{z'}(h_{xx}x' + h_{xy}y')$ and $\tilde F_{x'} = F_{x'} + F_{z'}h_x$, whence $\frac{\mathrm d}{\mathrm dt}\tilde F_{x'} = \frac{\mathrm d}{\mathrm dt}F_{x'} + h_x\frac{\mathrm d}{\mathrm dt}F_{z'} + F_{z'}(h_{xx}x' + h_{xy}y')$. Subtracting, $\tilde E_x = E_x + h_xE_z$, and likewise $\tilde E_y = E_y + h_yE_z$. Both vanish, so $E_x = (G_x/G_z)E_z$, $E_y = (G_y/G_z)E_z$, and trivially $E_z = (G_z/G_z)E_z$. That is $\mathbf E = (E_z/G_z)\nabla G$; put $\lambda = -E_z/G_z$, continuous because everything in it is. Near a point where another partial is non-zero, repeat with that variable; where two charts overlap the two formulas for $\lambda$ agree, since $\nabla G\ne0$ determines the coefficient uniquely. ∎

Why $\tfrac12\lvert\mathbf q'\rvert^2$ and not arc length. Its Beltrami quantity $F - \mathbf q'\cdot F_{\mathbf q'} = -\tfrac12\lvert\mathbf q'\rvert^2$ is conserved, so its extremals run at constant speed, and they avoid the reparametrisation degeneracy of the previous slide's notes. They are also the right curves: by Cauchy–Schwarz, $L^2 = \big(\int\lvert\mathbf q'\rvert\,\mathrm dt\big)^2 \le (t_1-t_0)\int\lvert\mathbf q'\rvert^2\,\mathrm dt$, with equality exactly at constant speed, so minimising the energy on a fixed time interval minimises the length too. Unit 32 defines geodesics this way.

The cross-product step. $(\mathbf q\times\mathbf q')' = \mathbf q'\times\mathbf q' + \mathbf q\times\mathbf q'' = \mathbf 0 + 2\lambda\,\mathbf q\times\mathbf q = \mathbf 0$ (unit 03's cross product, and $\mathbf a\times\mathbf a = \mathbf 0$). So $\mathbf N = \mathbf q\times\mathbf q'$ is constant, non-zero for a moving curve, and $\mathbf q\cdot\mathbf N = 0$ for all $t$: the curve lies in the plane through the centre with normal $\mathbf N$. A plane through the centre cuts the sphere in a great circle. Physically, this is a bead on a frictionless sphere with no gravity. Problems P8 and P9 take geodesics further: onto surfaces of revolution, and past the antipode.

11 · constraints

On the sphere, the multiplier is the normal force

and $\lambda$ is the normal force

Differentiating $\lvert\mathbf q\rvert^2 = R^2$ twice, $\mathbf q\cdot\mathbf q'' = -\lvert\mathbf q'\rvert^2$, so $2\lambda = -\lvert\mathbf q'\rvert^2/R^2$: the constraint supplies the centripetal acceleration $\lvert\mathbf q'\rvert^2/R$, inward. As with the chain, the multiplier is the force of constraint.

Unit 12 makes this the rule for every constrained system: a multiplier is the force that enforces its constraint.

The motion seen face-on, in the plane of the great circle. At speed $\lvert\mathbf q'\rvert$ round a circle of radius $R$ the acceleration points to the centre with size $\lvert\mathbf q'\rvert^2/R$ — and $\mathbf q'' = 2\lambda\mathbf q$ says exactly that, with $2\lambda R = -\lvert\mathbf q'\rvert^2/R$.

section 5

Stationary is not minimal

The second derivative, a condition on it that is necessary and not sufficient, and the conjugate points that close the gap.

11 · second variation

The second variation

definition

The second variation is $\ \delta^2J[y;\eta] = \varphi''(0)$ for $\varphi(\varepsilon) = J[y + \varepsilon\eta]$. Differentiating under the integral twice,

$$\delta^2J[y;\eta] = \int_a^b\Big(F_{yy}\,\eta^2 + 2F_{yy'}\,\eta\eta' + F_{y'y'}\,\eta'^2\Big)\mathrm dx .$$

necessary condition

If $y$ is a weak local minimum then $\delta^2J[y;\eta]\ge0$ for every $\eta\in C^1_0$.

Proof. $\varphi$ has a minimum at $0$ with $\varphi'(0) = 0$; Taylor's theorem in one variable forces $\varphi''(0)\ge0$. ∎

the form to remember

For $y\in C^2$, write $2\eta\eta' = (\eta^2)'$ and integrate that term by parts:

$$\delta^2J = \int_a^b\big(P\,\eta'^2 + Q\,\eta^2\big)\,\mathrm dx,$$

$P = F_{y'y'}$, $\ Q = F_{yy} - \dfrac{\mathrm d}{\mathrm dx}F_{yy'}$, all along $y$.

The integration by parts. Along a $C^2$ extremal with $F\in C^3$, the function $x\mapsto F_{yy'}(x,y,y')$ is $C^1$. Then $\int 2F_{yy'}\eta\eta' = \int F_{yy'}(\eta^2)' = [F_{yy'}\eta^2]_a^b - \int (F_{yy'})'\eta^2$, and the bracket vanishes at both ends. Collect terms.

What it is. $\delta^2J[y;\cdot]$ is a quadratic form on the vector space $C^1_0$, and it plays exactly the part the Hessian played in unit 04's classification of critical points. The difference is the one that makes this section necessary: in $\mathbb R^n$ a positive-definite Hessian gives a strict local minimum, because the unit sphere is compact and the form is bounded below on it by a positive constant. In $C^1_0$ neither holds automatically. A form can be positive on every non-zero $\eta$ and still have infimum $0$ on the unit sphere, and then higher-order terms can win. The sufficient condition needs strong positivity, $\delta^2J[\eta]\ge c\lVert\eta\rVert^2$ in a suitable norm — which is what Jacobi's theorem, later in this section, delivers. Cited: Gelfand and Fomin, chapter 5.

Dimensions. $[P\,\eta'^2]\cdot[\mathrm dx] = [F_{y'y'}]\,[y']^2[x] = [J]$, and likewise for $Q\eta^2$, so $\delta^2J$ has the dimension of $J$ when $\eta$ has the dimension of $y$ — as a second derivative in the dimensionless $\varepsilon$ should.

11 · second variation

The Legendre condition

theorem · Legendre condition

If $y\in C^2$ is a weak local minimum, then $P(x) = F_{y'y'}\big(x, y(x), y'(x)\big)\ge0$ on $[a,b]$.

proof

Suppose $P(x_0) < 0$ at some interior $x_0$ (the ends follow by continuity). Fix a non-zero $C^1$ function $\phi$ on $\mathbb R$ vanishing outside $(-1,1)$ and squeeze it: $\eta_w(x) = w\,\phi\big((x-x_0)/w\big)$. Then $\eta_w' = \phi'\big((x-x_0)/w\big)$ is $O(1)$ while $\eta_w = O(w)$, and substituting $x = x_0 + ws$,

$$\frac{\delta^2J[\eta_w]}{w} = \int_{-1}^{1}P(x_0 + ws)\,\phi'(s)^2\,\mathrm ds + w^2\!\int_{-1}^{1}Q(x_0+ws)\,\phi(s)^2\,\mathrm ds \;\longrightarrow\; P(x_0)\!\int\phi'^2 < 0 .$$

So $\delta^2J[\eta_w] < 0$ for small $w$, contradicting the necessary condition. ∎

On the cycloid. $P = \big(2gy\big)^{-1/2}(1+y'^2)^{-3/2} > 0$ everywhere on $(0, X]$ — at $B$ it is $0.2146\ \mathrm{s\,m^{-1}}$. It passes, strictly.

The reading. A variation that wiggles fast costs through $P\eta'^2$ long before it gains through $Q\eta^2$. So $P$ decides the sign of short wiggles and nothing else.

The limit, justified. $P$ and $Q$ are continuous on $[a,b]$, hence bounded and uniformly continuous, so $P(x_0 + ws)\to P(x_0)$ uniformly in $s\in[-1,1]$ as $w\to0$ and the first integral tends to $P(x_0)\int\phi'^2$, while the second is bounded and is multiplied by $w^2$. For $w$ small enough that $[x_0 - w, x_0 + w]\subset(a,b)$, $\eta_w$ is admissible. An explicit $\phi$: $\phi(s) = (1 - s^2)^2$ on $[-1,1]$, which is $C^1$ with $\phi(\pm1) = \phi'(\pm1) = 0$, extended by zero.

Endpoints. If $P(a) < 0$, continuity gives $P < 0$ at some interior point nearby, and the argument runs there.

Why only weak minima are needed. For each fixed $\eta$, the curves $y + \varepsilon\eta$ tend to $y$ in the $\lVert\cdot\rVert_1$ norm as $\varepsilon\to0$, so if $y$ is a weak local minimum then $\varphi$ has a local minimum at $0$. The squeezed bump is used only through $\varphi''(0)$.

The units of $P$ at $B$. $F$ is $\mathrm{s\,m^{-1}}$ and $y'$ is dimensionless, so $F_{y'y'}$ is $\mathrm{s\,m^{-1}}$. With $y = 1$ m and $y' = \cot(\theta_B/2) = -0.18547$ there, $1 + y'^2 = 1.03440$ — which is $2a/Y$, as the first integral says it must be — and $P = (2\times9.81\times1)^{-1/2}(1.03440)^{-3/2} = 0.2146\ \mathrm{s\,m^{-1}}$.

Where it goes. $P > 0$ everywhere, for every $y$ — convexity of $F$ in $y'$ — is the hypothesis that makes the Legendre transform to a Hamiltonian well defined (unit 14), and that makes the direct method of section 6 work. Legendre (1786) believed it was also sufficient for a minimum. The next slide is why it is not.

11 · second variation

Legendre is not enough

stationary, Legendre-positive, and not a minimum

$J_\ell[y] = \displaystyle\int_0^\ell\big(y'^2 - y^2\big)\,\mathrm dx$, $\ y(0) = y(\ell) = 0$. The curve $y\equiv0$ is an extremal (Euler–Lagrange reads $y'' + y = 0$), and $P = 2 > 0$. Now test $\eta = \sin(\pi x/\ell)$:

$$J_\ell[\varepsilon\eta] = \varepsilon^2\!\int_0^\ell\Big(\frac{\pi^2}{\ell^2}\cos^2\frac{\pi x}{\ell} - \sin^2\frac{\pi x}{\ell}\Big)\mathrm dx = \varepsilon^2\,\frac{\ell}{2}\Big(\frac{\pi^2}{\ell^2} - 1\Big).$$

For $\ell > \pi$ this is negative for every $\varepsilon\ne0$. So $y\equiv0$ is not even a weak local minimum: curves as close as you like, in slope as well as position, do better. At $\ell = 4$, $J_4[\eta] = -0.7663$.

11 · second variation

Half a period: where the minimum is lost

For $\ell < \pi$ the same curve is the minimum: $J_\ell[\eta] > 0$ for every non-zero $\eta$, not just the sine. The proof needs conjugate points and Jacobi's theorem, which come next.

The physics. With $x$ as time, $y'^2 - y^2$ is kinetic minus potential energy for an oscillator of angular frequency $1$. Its action is least for durations under half a period, and merely stationary beyond.

$J_\ell$ on the test variation $\sin(\pi x/\ell)$, against the length of the interval. The slope term $\pi^2/(2\ell)$ falls as the interval grows and the $-\ell/2$ term wins at $\ell = \pi$, half the oscillator's period.

Reading the curve. $J_\ell[\sin(\pi x/\ell)] = \pi^2/(2\ell) - \ell/2$: the first term is what the variation's slope costs, spread over a longer interval and so cheaper, the second what its height gains. They balance exactly at $\ell = \pi$. Positivity on this one $\eta$ for $\ell < \pi$ proves nothing about the others — that is what Jacobi's theorem supplies — but negativity for $\ell > \pi$ is already a complete proof that $y\equiv0$ is not a minimum there.

Where the threshold comes from. The Euler–Lagrange equation $y'' + y = 0$ has the solutions $c\sin x$ through the origin, and every one of them returns to zero at $x = \pi$. A family of extremals leaving $0$ together and meeting again at $\pi$: that is the phenomenon the next slide names. Before $\pi$ the family fans out without crossing, and the extremal is a minimum; after $\pi$ it is not.

Why this example is the right one to keep. It is the harmonic oscillator, with $m = 2$ and $\omega = 1$ to keep the numbers clean, and unit 12 will show that Newton's equations are the Euler–Lagrange equations of $\int(T - V)\,\mathrm dt$. So "the action is minimised" is false in general for the simplest system in physics, once the time interval exceeds half a period. The course says stationary action for this reason. On the sphere, the same phenomenon is a great-circle arc longer than half the circle (problem P9).

11 · second variation

The accessory equation and conjugate points

definitions

The accessory equation (Jacobi equation) of $J$ along $y$ is the Euler–Lagrange equation of the quadratic functional $\delta^2J = \int(Ph'^2 + Qh^2)$:

$$\big(P\,h'\big)' = Q\,h .$$

A point $c\in(a,b]$ is conjugate to $a$ along $y$ if some solution $h\not\equiv0$ has $h(a) = 0 = h(c)$.

theorem · where solutions come from

If $y(x;\alpha)$ is a smooth family of extremals, then the Jacobi field $h = \partial_\alpha y$ solves the accessory equation along each member.

Proof. Differentiate $F_y - (F_{y'})' = 0$ in $\alpha$: $F_{yy}h + F_{yy'}h' - \big(F_{yy'}h + F_{y'y'}h'\big)' = 0$. The $F_{yy'}h'$ terms cancel, leaving $(Ph')' = Qh$. ∎

The accessory equation, derived. For the integrand $\Phi(x,h,h') = Ph'^2 + Qh^2$, $\Phi_{h'} = 2Ph'$ and $\Phi_h = 2Qh$, so Euler–Lagrange is $(2Ph')' = 2Qh$. It is linear in $h$: the extremals of a quadratic functional form a vector space. The name "accessory" records the auxiliary role the problem of minimising $\delta^2J$ plays for the original one.

The family theorem, fully. Assume $F\in C^3$ and $y(x;\alpha)$ is $C^3$ in $(x,\alpha)$ jointly, so that $\alpha$-derivatives commute with $x$-derivatives (unit 03's Clairaut theorem). Along each member, $F_y(x,y,y') - \frac{\mathrm d}{\mathrm dx}F_{y'}(x,y,y') = 0$ identically in $(x,\alpha)$. Apply $\partial_\alpha$, using the chain rule and $\partial_\alpha y' = (\partial_\alpha y)' = h'$: $\partial_\alpha F_y = F_{yy}h + F_{yy'}h'$ and $\partial_\alpha F_{y'} = F_{y'y}h + F_{y'y'}h'$. So $F_{yy}h + F_{yy'}h' - (F_{yy'}h)' - (Ph')' = 0$, and $(F_{yy'}h)' = (F_{yy'})'h + F_{yy'}h'$. The $F_{yy'}h'$ terms cancel, leaving $\big(F_{yy} - (F_{yy'})'\big)h = (Ph')'$, which is $Qh = (Ph')'$.

11 · second variation

Conjugate points, drawn

so a conjugate point is

where extremals leaving $a$ together, infinitesimally apart, cross again. For $J_\ell$, $P = 2$ and $Q = -2$, so the accessory equation is $h'' + h = 0$; the family of extremals $\alpha\sin x$ gives $h = \sin x$, and the first conjugate point is $\pi$ — exactly the threshold found on the half-period slide.

The extremals $y = \alpha\sin x$ of $J_\ell$ through $0$. However they fan out, all of them return to the axis at $x = \pi$; an interval of length $4$ contains that refocusing point, and the zero extremal stops being a minimum.

11 · second variation

Jacobi's theorem

proved here

If $P > 0$ on $[a,b]$ and the accessory equation has a solution $u > 0$ on all of $[a,b]$, then $\delta^2J[\eta] > 0$ for every non-zero $\eta\in C^1_0$.

proof · complete the square (the Picone identity)

Using $(Pu')' = Qu$, one checks by differentiating the last term that

$$P\eta'^2 + Q\eta^2 = P\Big(\eta' - \frac{u'}{u}\eta\Big)^2 + \Big(\frac{P u'}{u}\,\eta^2\Big)' .$$

Integrate over $[a,b]$: the derivative term gives $0$ because $\eta(a)=\eta(b)=0$. What is left is $\ge 0$, and $= 0$ only if $(\eta/u)' = 0$, so $\eta = cu$ with $c = 0$ from $\eta(a) = 0$. ∎

cited · necessity

If $P > 0$ and some $c\in(a,b)$ is conjugate to $a$, then $\delta^2J$ takes negative values: not a minimum. Gelfand and Fomin, chapter 5. The proof needs uniqueness for linear ODEs (unit 01), which this unit does not assume.

cited · sufficiency

If $P > 0$ on $[a,b]$ and no point of $(a,b]$ is conjugate to $a$, then $y$ is a strict weak local minimum. Same reference.

Full proof of the identity

Write $w = u'/u$, which is $C^1$ because $u > 0$ is $C^2$. Expand the square: $P(\eta' - w\eta)^2 = P\eta'^2 - 2Pw\eta\eta' + Pw^2\eta^2$. Differentiate the other term: $(Pw\eta^2)' = (Pw)'\eta^2 + 2Pw\eta\eta'$. Now $Pw = Pu'/u$, so $(Pw)' = (Pu')'/u - Pu'^2/u^2 = Q - Pw^2$, using the accessory equation. Adding, $P\eta'^2 - 2Pw\eta\eta' + Pw^2\eta^2 + (Q - Pw^2)\eta^2 + 2Pw\eta\eta' = P\eta'^2 + Q\eta^2$. ∎

The equality case. $\int P(\eta' - w\eta)^2 = 0$ with $P > 0$ and a continuous integrand forces $\eta' = w\eta = (u'/u)\eta$, i.e. $(\eta/u)' = (\eta'u - \eta u')/u^2 = 0$. So $\eta/u$ is constant, and it is $0$ at $a$.

How it connects to conjugate points. If $u > 0$ on $[a,b]$ and $h$ is a solution with $h(a) = 0$, Sturm's comparison argument (which needs unit 01's uniqueness) shows $h$ cannot vanish again in $(a,b]$. Conversely, when there is no conjugate point in $(a,b]$, one takes the solution with $h(a - \delta) = 0$ for a small $\delta$ and shows it stays positive on $[a,b]$, which needs continuous dependence on initial data. Both directions are in Gelfand and Fomin, chapter 5, and are cited, not proved, here. In every example of this unit a positive solution $u$ is written down explicitly, so nothing depends on the cited parts.

On $J_\ell$ with $\ell < \pi$. $u(x) = \sin\big(x + (\pi - \ell)/2\big)$ solves $u'' + u = 0$, which is the accessory equation with $P = 2$, $Q = -2$, and its argument runs over $[(\pi-\ell)/2,\,(\pi+\ell)/2]\subset(0,\pi)$, so $u > 0$ on $[0,\ell]$. Hence $\int_0^\ell\eta'^2 > \int_0^\ell\eta^2$ for every non-zero $\eta\in C^1_0[0,\ell]$ — a form of Wirtinger's inequality, and the proof the "Legendre is not enough" slide promised.

And $\pi$ is the first conjugate point, with no ODE uniqueness needed. If some $h\not\equiv0$ solved $h'' + h = 0$ with $h(0) = 0 = h(c)$ and $0 < c < \pi$, then multiplying by $h$ and integrating by parts, $\int_0^c(h'^2 - h^2)\,\mathrm dx = [hh']_0^c = 0$ — contradicting the strict inequality just proved on $[0,c]$. (Here $h$ is not identically zero on $[0,c]$: $h'^2 + h^2$ has derivative $2h'(h''+h) = 0$, so it is constant, and it would be $0$ everywhere if $h$ vanished on an interval.)

11 · second variation · example

Worked: the cycloid has no conjugate point

a Jacobi field for the cycloid

Step 1 · a family. Substituting $x = \lambda s$ gives $T[\lambda\,y(\cdot/\lambda)] = \sqrt\lambda\,T[y]$, so dilating an extremal from $A$ gives another. The cycloids through $A$ are $y_a(x) = a\,f(x/a)$, with $f$ the cycloid of radius $1$.

Step 2 · differentiate in $a$ at fixed $x$, with $s = x/a$: $\ h = \partial_a y_a = f(s) - s\,f'(s)$.

Step 3 · in terms of $\theta$. $f = 1 - \cos\theta$, $s = \theta - \sin\theta$, $f' = \sin\theta/(1-\cos\theta) = \cot(\theta/2)$, so

$$h = \frac{2 - 2\cos\theta - \theta\sin\theta}{1-\cos\theta} = 2 - \theta\cot\tfrac\theta2 .$$

Step 4 · zeros. $h = 0$ exactly when $\tan(\theta/2) = \theta/2$. On $(0,\pi)$, $\tan u > u$ makes $h > 0$; on $[\pi,2\pi)$, $\cot(\theta/2)\le0$ makes $h\ge2$. The first zero after $A$ is at $\theta = 8.9868$, beyond the arch's end at $2\pi$.

Step 5 · verdict. $\theta_B = 3.5084$: no point of $(0, X]$ is conjugate to $A$, $h(\theta_B) = 2.651$, and $P > 0$. Jacobi's test passes.

checks

Dimension. $h = \partial y/\partial a$ is a pure number ✓.

Near $A$. $h\approx\theta^2/6 \to 0$: the field vanishes at $A$, as a Jacobi field from $A$ must.

Déjà vu. $h$ is the numerator of $r'(\theta)$ from the constants slide. "Exactly one cycloid through each $B$" and "no conjugate point" are one fact.

Step 1 in full. Let $y_\lambda(x) = \lambda y(x/\lambda)$. Then $y_\lambda'(x) = y'(x/\lambda)$, and with $x = \lambda s$, $T[y_\lambda] = \int\sqrt{(1 + y'(s)^2)/(2g\lambda y(s))}\;\lambda\,\mathrm ds = \sqrt\lambda\,T[y]$. The same substitution sends a variation $\eta$ of $y$ to the variation $\lambda\eta(\cdot/\lambda)$ of $y_\lambda$, so $\delta T[y_\lambda;\eta_\lambda] = \sqrt\lambda\,\delta T[y;\eta]$, and stationarity is preserved. This is unit 00's mechanical similarity once more.

Step 2. $\partial_a\big(af(x/a)\big) = f(x/a) + a f'(x/a)\cdot(-x/a^2) = f(s) - sf'(s)$. Step 3's algebra: $(1-\cos\theta)^2 + \sin^2\theta = 2 - 2\cos\theta$, and $\theta\sin\theta/(1-\cos\theta) = \theta\cot(\theta/2)$. Near $\theta = 0$, $\theta\cot(\theta/2) = 2 - \theta^2/6 + O(\theta^4)$, so $h\approx\theta^2/6$. The root of $\tan u = u$ in $(\pi, 3\pi/2)$ is $u = 4.49341$, giving $\theta = 8.98682$.

What this does and does not prove. $A$ is a singular point of the problem — $y = 0$ there and $F$ is infinite — so Jacobi's sufficiency theorem, stated for $F$ smooth up to the ends, does not apply literally. What is proved: on any $[x_\epsilon, X]$ with $x_\epsilon > 0$, $h > 0$ and $P > 0$, so the proved half of Jacobi's theorem gives $\delta^2T > 0$ for every non-zero variation supported away from $A$. The full statement — that the cycloid is the global minimum among graphs — needs Weierstrass's field, the next slide.

A contrast worth seeing. Had the endpoint needed $\theta_B > 8.9868$ — impossible here, since the arch ends at $2\pi$ — the extremal would have run past a conjugate point. Problem P9 shows the same computation on the sphere, where the conjugate point is the antipode and does lie on the extremal.

11 · second variation

The cycloids through $A$ form a field

field of extremals

A family of extremals that covers a region exactly once: through each point of it passes one member.

The first arches through $A$ do this for the whole quadrant $x, y > 0$: a point $(x,y)$ lies on the arch with $r(\theta) = x/y$, and $r$ is a bijection (the constants slide).

cited · Weierstrass

An extremal embedded in a field, with $F$ convex in $y'$ throughout, beats every admissible curve lying in the field's region. Gelfand and Fomin, chapter 6.

The arches leave $A$ together and never meet again — the Jacobi field $h = \partial_a y_a$ is their spacing, and it has no zero. Every admissible wire lies in the quadrant, so the cycloid to $B$ beats them all.

Why it applies to the bead. The integrand $F = \sqrt{1+p^2}/\sqrt{2gy}$ has $F_{pp} = (1+p^2)^{-3/2}/\sqrt{2gy} > 0$ for every slope $p$, not only along the cycloid, so it is convex in $y'$ everywhere in the quadrant. Every admissible wire has $y > 0$ on $(0, X]$, so it lies in the region the field covers. That is the whole of Weierstrass's hypothesis, and his theorem then makes the cycloid the minimum among all graphs from $A$ to $B$ — the strong, global conclusion that Jacobi's second-variation test, being local and in the weak norm, cannot reach.

Why the field exists is this section's computation seen from another side: the first arches of all the cycloids through $A$ cover the open quadrant exactly once because $r$ is strictly increasing, and $r' > 0$ is $h > 0$ is "neighbouring cycloids do not cross". What Weierstrass adds, and what is cited rather than proved here, is the argument that a field turns the comparison of two whole curves into an integral of a pointwise non-negative quantity — his excess function — along the competitor.

section 6

Does a minimum exist at all?

Everything so far assumed a minimiser and asked what it satisfies. Weierstrass showed the assumption can fail.

11 · existence

The direct method

In $\mathbb R^n$ (unit 04), a continuous function on a closed bounded set attains its minimum, because such sets are compact. The direct method tries the same thing on curves:

three steps

1. Take a minimising sequence: admissible $y_n$ with $J[y_n]\to\inf J$.

2 · compactness. Extract a subsequence converging, in some sense, to some $y_*$.

3 · lower semicontinuity. Show $J[y_*]\le\liminf J[y_n]$, and that $y_*$ is admissible. Then $J[y_*] = \inf J$.

cited · Tonelli's theorem

If $F$ is continuous, convex in $y'$, and $F\ge\alpha\lvert y'\rvert^q - \beta$ with $\alpha > 0$, $q > 1$, then $J$ attains its minimum over absolutely continuous curves with the given ends. Dacorogna, Introduction to the Calculus of Variations, chapter 3.

What Tonelli's theorem is and is not doing here. It is cited, because its proof needs Arzelà–Ascoli and weak convergence in $L^q$, neither of which this course develops. Its conclusion is a minimiser in a space larger than $C^1$; a separate regularity theorem (also cited, same reference) then brings it back to $C^1$, where section 2 applies. The brachistochrone does not satisfy its growth hypothesis — $F$ grows like $\lvert y'\rvert$, so $q = 1$ — and its existence comes instead from Weierstrass's field argument, cited on The cycloids through $A$ form a field. So for the running example existence and minimality arrive together.

Why a physicist should care. Every variational principle in this course assumes that nature's path is the minimiser or at least a stationary point, and derives equations from that. The direct method is the discipline of checking that the object exists before describing it — and Weierstrass against Dirichlet is what happened the one famous time nobody checked.

11 · existence

Bounded slopes, lost in the limit

the tension between steps 2 and 3

A weaker notion of convergence makes step 2 easier and step 3 harder. In $C^1$ with $\lVert\cdot\rVert_1$ the unit ball is not compact: $\eta_n = \sin(n\pi x)/(n\pi)$ has $\lVert\eta_n\rVert_1\le 2$ and no convergent subsequence.

Positions converge, to $0$, uniformly. Slopes do not converge at all. So a limit taken in the only sense available forgets the slopes — and $J$ depends on them.

$\eta_n = \sin(n\pi x)/(n\pi)$ above, its slope $\cos(n\pi x)$ below. Raising $n$ flattens the curve but leaves the slope swinging between $\pm1$, ever faster: bounded in $\lVert\cdot\rVert_1$, with nothing to converge to.

The non-compact sequence. $\eta_n' = \cos(n\pi x)$, so $\lVert\eta_n\rVert_1 = 1/(n\pi) + 1 \le 2$. A subsequence converging in $\lVert\cdot\rVert_1$ would have its slopes $\cos(n\pi x)$ converging uniformly, hence forming a Cauchy sequence in the mean square. But $\int_0^1(\cos n\pi x - \cos m\pi x)^2\,\mathrm dx = \tfrac12 + \tfrac12 = 1$ whenever $m \ne n$, so no subsequence is Cauchy even in that weaker sense. The positions $\eta_n$ do converge uniformly, to $0$. That is the typical picture: bounded slopes buy convergence of positions but not of slopes, and step 3 must then survive a limit in which slopes are lost. Convexity of $F$ in $y'$ is exactly what makes $J$ lower semicontinuous under that kind of convergence — the Legendre condition, imposed everywhere.

11 · existence

Weierstrass against Dirichlet

a functional with no minimiser

$J[y] = \displaystyle\int_{-1}^{1}x^2\,y'^2\,\mathrm dx$, $\quad y(-1) = -1$, $y(1) = 1$.

The infimum is $0$. Try $y_\epsilon = \arctan(x/\epsilon)/\arctan(1/\epsilon)$. With $A = \arctan(1/\epsilon)$, $y_\epsilon' = \epsilon/\big(A(x^2+\epsilon^2)\big)$, and $x^2\le x^2+\epsilon^2$ gives

$$J[y_\epsilon]\le\frac{\epsilon^2}{A^2}\int_{-1}^{1}\frac{\mathrm dx}{x^2+\epsilon^2} = \frac{2\epsilon}{A}\;\longrightarrow\;0 .$$

It is not attained. $J[y] = 0$ forces $xy' \equiv 0$, so $y$ is constant on $[-1,0)$ and on $(0,1]$ — hence $-1$ and $+1$ there. No continuous $y$ does that.

The minimising sequence steepens into a jump. At $\epsilon = 0.01$ the bound is $0.012814$ and the true value $0.006366$; the limit is not continuous, so not admissible.

The computation. $\frac{\mathrm d}{\mathrm dx}\arctan(x/\epsilon) = \epsilon/(x^2+\epsilon^2)$, and $y_\epsilon(\pm1) = \pm1$ because $\arctan$ is odd. Then $x^2y_\epsilon'^2 = \epsilon^2x^2/\big(A^2(x^2+\epsilon^2)^2\big) \le \epsilon^2/\big(A^2(x^2+\epsilon^2)\big)$, and $\int_{-1}^{1}\mathrm dx/(x^2+\epsilon^2) = (2/\epsilon)\arctan(1/\epsilon) = 2A/\epsilon$. So $J[y_\epsilon]\le 2\epsilon/A$, and $A\to\pi/2$, so the bound behaves like $4\epsilon/\pi$. At $\epsilon = 0.01$, $A = 1.56080$ and $2\epsilon/A = 0.012814$; direct quadrature gives $0.006366$.

The non-attainment, carefully. If $y\in C^1$ has $J[y] = 0$, the continuous non-negative integrand $x^2y'^2$ vanishes identically (the lemma in section 2's notes), so $y' = 0$ wherever $x \ne 0$. On $[-1,0)$ the mean value theorem makes $y$ constant, equal to $y(-1) = -1$; on $(0,1]$ it equals $1$. Continuity at $0$ is then impossible. So no admissible curve reaches the infimum, and the Euler–Lagrange equation $(x^2y')' = 0$ — whose solutions $y = k - C/x$ are singular at $0$ unless $C=0$, and then constant — has no admissible solution either.

11 · existence

What failed: a functional blind to steepness

what failed

$P = F_{y'y'} = 2x^2$ vanishes at $x = 0$: there the functional charges nothing for steepness, so the growth hypothesis of Tonelli fails, and the sequence escapes the space.

the casualty

Riemann had assumed, as Dirichlet's principle, that the energy $\int\lvert\nabla u\rvert^2$ always attains its minimum among functions with given boundary values; this example (1870) broke the argument.

Where the cost goes, for $\epsilon = 0.03$. The slope $y_\epsilon'$ spikes at $x = 0$, exactly where the weight $x^2$ vanishes. The integrand never exceeds $1/(4A^2) \approx 0.105$, however steep the slope, while its width shrinks with $\epsilon$ — so its area $J$ goes to $0$.

Reading the figure. $J$ is the area under the integrand, shaded (here scaled to peak $1$). All of the sequence's steepness is spent in a window of width about $\epsilon$ around the origin, and there the weight is about $\epsilon^2$; the slope there is about $1/\epsilon$, so the integrand is of order $1$ on a window of width $\epsilon$ and the area is of order $\epsilon$ — the $2\epsilon/A$ bound of the previous slide. A functional with $P$ bounded below by a positive constant would charge $1/\epsilon^2$ for that slope and the area would blow up instead: that is what Tonelli's growth hypothesis buys. Exactly: $x^2y_\epsilon'^2 = \epsilon^2x^2/\big(A^2(x^2+\epsilon^2)^2\big)$ is largest at $x = \pm\epsilon$, where it equals $1/(4A^2)$, which is $0.1053$ at $\epsilon = 0.03$.

The history. Dirichlet's principle — solve Laplace's equation (unit 23) by minimising the energy $\int\lvert\nabla u\rvert^2$ over functions with given boundary values — was used by Gauss, Kelvin and Dirichlet, and named and relied upon by Riemann in 1851 and 1857, including for his mapping theorem. Weierstrass presented this one-dimensional counterexample to the Berlin Academy in 1870: a positive functional need not attain its infimum. For thirty years the principle was suspect, and mathematicians such as Schwarz and Neumann found other routes to Riemann's results. Hilbert rehabilitated it in 1900, by proving existence for the Dirichlet energy under proper hypotheses — the first use of what is now the direct method. The episode is why the existence question is part of this unit.

11 · close

The bead, closed

the running example, start to finish

1. Written as a functional, $T[y] = \int\sqrt{(1+y'^2)/(2gy)}\,\mathrm dx$, with no mass in it.

2. Beltrami gave $y(1+y'^2) = 2a$ — Snell's law for a bead.

3. Integrated to the cycloid; $B$ fixed $\theta_B = 3.50837$, $a = 0.51720$ m, $T = 0.80556$ s, beating the line ($1.00964$ s) and Galileo's arc ($0.83031$ s).

4. Freed at the end, it lands horizontally: $0.80030$ s to the line $x = 2$ m.

5. Legendre $P > 0$; the dilation field $h = 2 - \theta\cot(\theta/2)$ has no zero before $\theta = 8.9868$; the cycloids through $A$ form a field of extremals, so Weierstrass's theorem (cited) makes it the quickest wire among graphs. Minimum, and it exists.

What this unit hands on

12 — Newton's equations are the Euler–Lagrange equations of $\int(T-V)\,\mathrm dt$, in any coordinates.

13 — the Beltrami identity and the first integral $F_{y'}$ become one theorem: symmetry gives conservation.

15 — the Snell reading of the cycloid is no accident: mechanics is optics.

26, 32 — proper time is maximised, and geodesics stop minimising at conjugate points: section 5, in spacetime.

Each step above is a slide or section of this deck, and each number was computed rather than quoted: $\theta_B$ as the unique root of $r(\theta) = 2$, $a$ from $Y/(1-\cos\theta_B)$, the times from $\theta\sqrt{a/g}$, the closed-form line time and a quadrature along the arc. The minimality claim has two parts of different status. The Legendre and Jacobi checks are proved on the slides. The field-of-extremals conclusion — global minimum among graphs — is cited (Gelfand and Fomin, chapter 6), with the fact it rests on, that the cycloids through $A$ never cross, proved on the constants slide.

Reading. Gelfand and Fomin, Calculus of Variations, chapters 1, 2 and 5 cover this unit's ground at the same level of rigour; chapter 4 is Hamiltonian and can wait for unit 14, and chapter 3's §§13–14 use variable endpoints in a way this unit only touches. One caution: chapter 5's proofs of the two cited halves of Jacobi's theorem lean on existence and uniqueness for linear ODEs, which is unit 01's Picard–Lindelöf and not a prerequisite here; read those proofs after unit 01, and the rest of the chapter now. For the history, Goldstine, A History of the Calculus of Variations from the 17th through the 19th Century, whose early chapters cover the Bernoullis, Euler and Lagrange.

11 · checkpoint

Done when

checkpoint

You can, on paper and without the deck,

derive the cycloid from scratch: write down the descent-time functional, reduce it by the Beltrami identity to a first-order equation, integrate that to $x = a(\theta - \sin\theta)$, $y = a(1-\cos\theta)$, and fix both constants from the endpoints; and
produce a functional whose Euler–Lagrange solution is stationary but not a minimum, together with the test that detects it — a conjugate point inside the interval, or an explicit variation making the second variation negative.

Falsifiable: a derivation that assumes the cycloid, or loses a constant, fails the first; "it satisfies Euler–Lagrange, so it is a minimum" fails the second. Record it with ./mr done 11.

Problems

./mr problems 11 — 11 problems, about 6 hours. Hints and worked solutions fold under each problem on the same page.

12 Lagrangian mechanics · 13 Noether · 15 Hamilton–Jacobi, which needs this unit directly · 19 Sturm–Liouville, where the accessory equation returns as an eigenvalue problem.

Try the checkpoint cold. For the first half, set the endpoint at $B = (1\ \mathrm m, 1\ \mathrm m)$ and check your constants against problem P2's: $\theta_B = 2.41201$ and $a = 0.57292$ m. For the second, the quickest honest answers are $\int_0^\ell(y'^2 - y^2)\,\mathrm dx$ with $\ell > \pi$, where $\sin(\pi x/\ell)$ makes the second variation negative and $\pi$ is the conjugate point, or a great-circle arc longer than half the circle, where the antipode is (problem P9).

If the first half stalls, it is usually at the half-angle substitution: redo the worked integration slide with $y = 2a\sin^2(\theta/2)$ from the start. If the second stalls, the usual confusion is between the Legendre condition, which the counterexample passes, and the Jacobi condition, which it fails.

The brachistochrone, and how to differentiate with respect to a curve

One question, asked of a curve

Functionals, and what it means to vary one

Three problems with one shape

Worked: Snell's law, from Fermat

Snell's law: least, not just stationary

The bead problem, written down

The bead's functional: no mass, and a singular start

The space the unknown lives in

The first variation is a directional derivative

Which minimum? Two norms, two answers

Close in position, far in slope

From "for every $\eta$" to one equation

The tool: a bump

The fundamental lemma of the calculus of variations

The du Bois-Reymond lemma

The Euler–Lagrange equation

Worked: the shortest path, and what the derivation assumed

The Beltrami identity: when $x$ is absent

Solving it: the cycloid

Beltrami on the bead

The first integral is Snell's law in layers

Worked: integrating the first integral

Why the answer is called a cycloid

The constants, and the race

Checking the answer from both ends

Constraints and free ends

A free end: the natural boundary condition

A free end, on the bead

An integral constraint: a multiplier in function space

Worked: the hanging chain

The multiplier is the tension

The tension, checked, and Galileo's parabola priced

Several unknowns, and higher derivatives

A pointwise constraint: curves on a surface

On the sphere, the multiplier is the normal force

Stationary is not minimal

The second variation

The Legendre condition

Legendre is not enough

Half a period: where the minimum is lost

The accessory equation and conjugate points

Conjugate points, drawn

Jacobi's theorem

Worked: the cycloid has no conjugate point

The cycloids through $A$ form a field

Does a minimum exist at all?

The direct method

Bounded slopes, lost in the limit

Weierstrass against Dirichlet

What failed: a functional blind to steepness

The bead, closed

What this unit hands on

Done when

Problems

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