Part A · the equation of motion · spine

Determinism, and the one hypothesis it rests on

Newton's laws determine the future — as a theorem, with a hypothesis, and a frictionless dome on which that hypothesis fails and the future does not.

needs 00~2 h slides + 5 h problemsproblems: ./mr problems 01

Unit 00 solved equations of motion in the three cases where a formula exists. This unit asks the prior question: when is there a solution at all, and when is there only one? The answer is the Picard–Lindelöf theorem, and it is the licence for every later unit that says the solution rather than a solution.

How to read this deck

One object runs through it: Norton's dome, a frictionless surface on which a mass sitting motionless at the apex can start to slide at a time of its own choosing, with no force applied and no law broken. It is introduced in section 1, it passes the theorem's test everywhere except at one point in section 2, it is taken apart in section 3, and section 4 says exactly how long its solutions live.

Words with a dotted underline are glossary terms; click one, or press G for the index. N opens the notes, where the long proofs are, under a Full proof heading. A solid thin underline names a piece of the running example and shows it on hover.

Nothing here needs vector calculus. Everything is single-variable calculus, linear algebra and metric-space topology, which is why this unit can come second.

01 · overview

One theorem, one hypothesis, one counterexample

1 · the claim

"The present state and the laws fix the future" is a statement about an initial value problem: it asserts existence and uniqueness.

2 · the theorem

Picard–Lindelöf: continuity plus a Lipschitz condition gives exactly one solution, for a while. Proved by a contraction on a space of curves.

3 · the counterexample

Norton's dome: continuity holds, Lipschitz fails at one point, and one initial condition has infinitely many futures: $r = k^2t^4/144$ solves $\ddot r = k\sqrt r$ from rest, and so does every delayed copy of it.

4 · the reach

Grönwall's inequality bounds how fast two futures separate; the maximal interval of existence says when a solution simply stops.

what this unit licenses

After it you may write the solution, verify a formula by substitution and call it the answer, and say precisely what a physicist means by determinism. Before it, none of those are available.

The order is deliberate. The theorem comes before the counterexample because the counterexample is only interesting once you know which hypothesis it breaks — otherwise it is a curiosity, and the usual reaction ("that cannot be physical") is a way of not looking at it.

Section 4 is the part that gets used most often later. Continuous dependence is what makes a numerical or approximate solution mean anything; the maximal interval is what unit 08's orbits and unit 37's infalling geodesics run into.

Where each box is done. Box 1 is Laplace, made precise. Box 2 is Picard–Lindelöf, proved on the two Proof slides after it, with the contraction coming from The contraction mapping principle. Box 3's formula is found by trying a power $r = \alpha t^p$ (the notes to The one-parameter family, verified) and checked on that slide; the delayed copies work because the equation does not contain $t$ and the glue at the departure time is $C^2$. Box 4 is Grönwall's inequality and The maximal interval of existence.

Problem sheet: ./mr problems 01. Five hours, and problems P10 and P11 are the ones the checkpoint is built from.

section 1

What the claim actually says

From a sentence about the universe to a statement about an ordinary differential equation.

01 · the claim

Laplace, made precise

Laplace, 1814: an intelligence that knew every force and "the respective situation of the beings who compose it" could compute the future and the past; "nothing would be uncertain, and the future, like the past, would be present to its eyes."

determinism, as mathematics

Given a law $\dot y = f(y, t)$ and a state $y_0$ at time $t_0$, there is exactly one function $y$ with $y(t_0) = y_0$ solving the equation.

Three separable assertions: a solution exists; it is unique; and it exists for all time. This unit proves the first two locally, and shows all three can fail.

what determinism is not

Not predictability. Unique solutions can separate exponentially, so a state known to 10 decimal places may be useless after a minute (unit 17).

Not a law of physics. It is a property some differential equations have and others do not, and which one you get is decided by the force law you wrote down.

Laplace's own formulation was a claim about a physical theory. Turning it into a theorem is the only way to find out that it is almost true, and to locate the exception.

The quotation is from the Essai philosophique sur les probabilités (1814), the introduction Laplace wrote for the second edition of his Théorie analytique. He states it to motivate probability: probability is what an intelligence less capable than the demon is reduced to. The demon is not the point of the passage; it is a foil.

Why "exactly one" and not "at least one". Existence alone gives a universe in which the future is possible, not one in which it is fixed. Uniqueness is the whole content of the claim. This is why section 3's counterexample — which has existence in abundance — is a counterexample to determinism.

The third assertion is genuinely separate. Section 4 exhibits a perfectly smooth law whose solution ceases to exist after a finite time. Nothing is singular about the law; the solution simply runs off to infinity, and there is no state to speak of afterwards.

A fourth assumption hides in the word "state": that the present is captured by finitely many numbers. For point particles that is what unit 00's second law gives us. Field theories (unit 18 onwards) replace $\mathbb{R}^{N}$ with a function space, and the theorems get harder in exactly the way you would expect.

01 · the claim

The state is a point of phase space

Position alone is not a state. For $\ddot x = -\omega^2 x$, the datum $x(0) = 0$ is satisfied by $x(t) = A \sin \omega t$ for every $A$: one position, a continuum of futures. Add $\dot x(0)$ and the ambiguity is gone.

phase space

For a system with configuration $x \in \mathbb{R}^n$, the state is the pair $y = (x, \dot x) \in \mathbb{R}^{2n}$, and $\mathbb{R}^{2n}$ is its phase space.

A point of phase space, not of space, is what a law of motion propagates.

Left: one position at $t = 0$, three futures. Right: the same three as states $(x, \dot x)$ — three different starting points on the $\dot x$-axis, each on its own circle.

The example, checked. $x(t) = A\sin\omega t$ has $\ddot x = -A\omega^2 \sin \omega t = -\omega^2 x$ and $x(0) = 0$ for every $A$. So the initial-value problem "position zero at time zero" has a one-parameter family of solutions — and this is not a failure of determinism, it is a failure to specify the state. Contrast section 3, where the full state is given and the family persists.

Reading the figure. On the left, three solutions of $\ddot x = -x$ (dimensionless, $\omega = 1$) with amplitudes $1$, $0.5$ and $-0.75$: they share the point $x(0) = 0$ and nothing else. On the right the same three are drawn as $t \mapsto (x(t), \dot x(t)) = (A\sin t, A\cos t)$, circles of radius $|A|$ traced clockwise. Their starting points $(0, A)$ are three different points of the plane, so in phase space there is no ambiguity left to resolve.

Phase space is where the whole of Part A lives from here: unit 02 draws pictures in it, unit 14 discovers it has a geometry of its own, and unit 17 measures how fast neighbouring points in it separate.

01 · the claim

Why the state is a pair

why the second law has this shape

Newton's law fixes $\ddot x$, so it is the pair $(x, \dot x)$ that must be supplied and the pair that is then determined. Had forces fixed the third derivative $x^{(3)}$, the state would be a triple and nature would need one more number per particle.

That the state is $2n$ numbers is a physical fact expressed as the order of an ODE. Unit 14 makes phase space the primary object; here it is a bookkeeping device that will not stay one.

Dimensions: the two halves of $y$ carry different units, $\mathrm{L}$ and $\mathrm{L\,T^{-1}}$. Any norm on $\mathbb{R}^{2n}$ therefore compares unlike quantities — nondimensionalize first, as unit 00 insisted, and every constant below is a pure number.

The law, as an arrow at each state. At $y = (x, v)$ the arrow $f(y)$ has horizontal part $v$ — the height of $y$, kinematics — and vertical part $F/m$, the force. The pair fixes the arrow; drawn for $\ddot x = -x$, whose solutions go round circles.

What the figure is saying. Read the second law as a rule that attaches an arrow to every point of phase space: at the state $y = (x, v)$ the arrow is $f(y) = (v, F/m)$, the velocity with which the state moves. Its horizontal part is the point's own height $v$ — that is the definition of velocity, true for every mechanical system — and only its vertical part carries the physics. A solution is a curve that is tangent to the arrow at every point. The next slide makes this precise for any $n$; unit 02 is the study of such pictures.

The units warning is not pedantry. The Lipschitz constants in this unit are suprema of quantities like $\|f(y)-f(z)\|/\|y-z\|$, and if the components of $y$ have different dimensions that ratio has no dimension at all. The fix is the one unit 00 taught: choose a length $\ell$ and a time $\tau$ intrinsic to the problem, put $\hat x = x/\ell$ and $\hat t = t/\tau$, and work with the dimensionless system. The pendulum becomes $\ddot\theta = -\sin\theta$ with $\tau = \sqrt{\ell/g}$, and its Lipschitz constant is the pure number 1.

01 · the claim

Every Newtonian law is a first-order system

theorem · phase-space reduction

Let $F : \mathbb{R}^n \times \mathbb{R}^n \times \mathbb{R} \to \mathbb{R}^n$ and $m > 0$. Define $f : \mathbb{R}^{2n} \times \mathbb{R} \to \mathbb{R}^{2n}$ by $f\big((x,v), t\big) = \big(v,\ \tfrac1m F(x, v, t)\big)$. Then $x$ is a $C^2$ solution of $m\ddot x = F(x, \dot x, t)$ on an interval $I$ if and only if $y = (x, \dot x)$ is a $C^1$ solution of $\dot y = f(y, t)$ on $I$.

proof

($\Rightarrow$) If $x$ is $C^2$ then $y = (x,\dot x)$ is $C^1$ and $\dot y = (\dot x, \ddot x) = \big(\dot x, \tfrac1m F(x,\dot x,t)\big) = f(y,t)$.

($\Leftarrow$) If $y = (y_1, y_2)$ is $C^1$ with $\dot y = f(y,t)$, the first block says $\dot y_1 = y_2$, so $y_1$ is $C^2$ with $\ddot y_1 = \dot y_2$; the second block says $\dot y_2 = \tfrac1m F(y_1, y_2, t)$. Hence $m \ddot y_1 = F(y_1, \dot y_1, t)$. $\square$

The theorem is trivial and it is the reason the rest of the unit is possible: one theorem about first-order systems $\dot y = f(y,t)$ covers every mechanical law, at every $n$, at once.

check

Dimensions: the two blocks of $f$ have units $\mathrm{L\,T^{-1}}$ and $\mathrm{L\,T^{-2}}$, matching $\dot x$ and $\ddot x$. Special case $F = F(t)$: the system is $\dot y_1 = y_2$, $\dot y_2 = F(t)/m$, and integrating twice is unit 00's first quadrature case.

The same trick reduces any order: $x^{(k)} = G(x, \dot x, \dots, x^{(k-1)}, t)$ becomes a first-order system on $\mathbb{R}^{kn}$ by taking the first $k-1$ derivatives as extra coordinates. And a non-autonomous system is made autonomous by adding one more coordinate with $\dot t = 1$, on $\mathbb{R}^{N+1}$ — useful in unit 02, where the pictures need a field that does not move.

What is lost. Nothing, mathematically. What is lost is the structure of the second-order equation: the split into "kinematics" ($\dot x = v$, always true) and "dynamics" ($m \dot v = F$, the physics). Unit 14 puts a different, better structure back in, in which the two blocks of $f$ are derivatives of a single function.

Why $m > 0$ matters. With $m = 0$ the second block is not solvable for $\dot v$, and the equation is not an ODE at all but a constraint. Massless particles are not a limit of this formalism; unit 27 treats them separately, and their trajectories come from a different equation.

01 · the running example

Norton's dome: the surface, and its equation

A frictionless surface of revolution, sitting in uniform gravity $g$. Let $r$ be the arc length from the apex down a meridian, and $h(r)$ the height, so that $h(0) = 0$ and $h \le 0$.

the equation of motion

With $r$ an arc length, the unit tangent to the meridian has vertical component $h'(r)$, so the tangential component of the weight $(0,0,-mg)$ is $-mg\,h'(r)$, and the normal force contributes nothing to it:

$$m\,\ddot r = -m g\, h'(r).$$

Norton's choice $\;h(r) = -\dfrac{2k}{3g} r^{3/2}\;$ gives $h'(r) = -\dfrac{k}{g}\sqrt{r}$ and hence

$$\boxed{\ \ddot r = k \sqrt{r\,}\ }$$

with $k > 0$ a constant of dimension $\mathrm{L^{1/2}\,T^{-2}}$. Take $k = 1\,\mathrm{m^{1/2}s^{-2}}$ throughout.

Schematic, curvature exaggerated. The arc length $r$ runs from the apex along the surface. Of the weight, only the part along the surface, $-mg\,h'(r)$, drives the motion; $N$ balances the rest.

Reading the figure. The weight $mg$ points straight down; split it along the surface and across it. The part across the surface is cancelled by the normal force $N$, which can do nothing else, since it is perpendicular to every possible motion. The part along the surface is $-mg\,h'(r)$, and it is the only force in the equation of motion. At the apex the tangent plane is horizontal, so this part is exactly zero there.

Why there is no $\sqrt{1 + h'^2}$ anywhere. Because $r$ is arc length, not horizontal radius. Writing the meridian as $(\rho(r), h(r))$ with $\rho'^2 + h'^2 = 1$, the unit tangent is $(\rho', h')$, and the component of $(0,-mg)$ along it is $-mg\,h'$ exactly. Parametrizing by $\rho$ instead would put the usual $1/\sqrt{1+({\rm d}h/{\rm d}\rho)^2}$ back.

Full derivation, the energy route

For motion down a meridian the speed is $|\dot r|$, so unit 00's first integral is $E = \tfrac12 m \dot r^2 + m g h(r)$. Differentiating, $m\dot r \ddot r + m g h'(r) \dot r = 0$, and at any instant with $\dot r \ne 0$ this gives $\ddot r = -g h'(r)$. Both sides are continuous, so the equation holds at the isolated instants where $\dot r = 0$ as well. This is exactly unit 00's third quadrature case (force depending on position) run backwards.

Norton introduced the dome

Norton introduced the dome in "Causation as folk science" (2003) and analysed it in "The dome: an unexpectedly simple failure of determinism", Philosophy of Science 75 (2008). The mathematics was old; the point he made with it was not.

01 · the running example

Norton's dome: where the surface ends

the surface must exist

A graph over arc length needs $|h'| \le 1$, i.e. $\dfrac{k}{g}\sqrt r \le 1$, i.e. $r \le g^2/k^2 = 96.17$ m. At that radius the meridian is vertical and the dome stops: it is $64.11$ m deep, and its rim is $64.11$ m from the axis.

dimensions

Norton writes $h = -\tfrac{2}{3g} r^{3/2}$, which is not a length. The missing $k$ is set to 1 in SI units. Keep it: it is what makes $r \le g^2/k^2$ readable, and what the problem sheet varies.

To scale, $k = 1\,\mathrm{m^{1/2}s^{-2}}$: as deep as it is wide in radius, with a vertical wall at the rim. The marked point is where the sliding mass leaves the surface, the subject of "How long does a dome solution live?".

Drawing it to scale. The meridian is $\big(\rho(r), h(r)\big)$ with $\rho' = \sqrt{1 - h'^2} = \sqrt{1 - (k/g)^2 r}$, so $\rho(r) = \tfrac{2g^2}{3k^2}\big(1 - (1 - k^2 r/g^2)^{3/2}\big)$; the figure plots this against $h(r) = -\tfrac{2k}{3g}r^{3/2}$ for $0 \le r \le g^2/k^2$, with $g = 9.80665\ \mathrm{m\,s^{-2}}$. Near the apex the dome is nearly flat — at $r = 1$ m it has dropped only $6.8$ cm — which is why it looks harmless.

Does the mass stay on the dome? Write $u = (k/g)\sqrt r = |h'|$. The normal force along the spontaneous solution is $N = mg\big(\sqrt{1-u^2} - \tfrac{2}{3}u^2/\sqrt{1-u^2}\big)$, derived below, which vanishes at $u^2 = 3/5$, that is at $r = 57.70$ m, where the surface is inclined at $50.77^\circ$. Beyond that the mass flies off. Nothing mathematical turns on this — the indeterminacy is at $r = 0$, where $N = mg$ — but a counterexample should not also be physically impossible, and this one is not.

Full derivation of the normal force

The meridian is $w(r) = (\rho(r), h(r))$ with $\rho'^2 + h'^2 = 1$, so $T = (\rho',h')$ is the unit tangent and $n = (-h', \rho')$ a unit normal, pointing away from the surface. Differentiating the position $w(r(t))$ twice by the chain rule,

$$\frac{\mathrm{d}^2}{\mathrm{d}t^2}\,w(r(t)) = (\rho',h')\,\ddot r + (\rho'',h'')\,\dot r^2 ,$$

and Newton's law reads $m\big[(\rho',h')\ddot r + (\rho'',h'')\dot r^2\big] = (0,-mg) + N\,n$. Take the inner product with $n$: the $\ddot r$ term drops out because $T \cdot n = 0$, and $(0,-mg)\cdot n = -mg\rho'$, so $N = mg\,\rho' + m\dot r^2\,\big[(\rho'',h'')\cdot n\big]$. (The tangential component is the equation of motion $m\ddot r = -mgh'$ again, which is how the two were separated in the first place.) Now $h' = -u$, $\rho' = \sqrt{1-u^2}$ and $\mathrm{d}u/\mathrm{d}r = k^2/(2g^2u)$, so $h'' = -\tfrac{k^2}{2g^2u}$, $\ \rho'' = -\tfrac{k^2}{2g^2\sqrt{1-u^2}}$, and

$$(\rho'',h'')\cdot n = \rho'' u + h''\rho' = -\frac{k^2}{2g^2}\Big(\frac{u}{\sqrt{1-u^2}} + \frac{\sqrt{1-u^2}}{u}\Big) = -\frac{k^2}{2g^2\,u\sqrt{1-u^2}} .$$

Along the family $\dot r^2 = \tfrac43 k r^{3/2} = \tfrac43 (g^3/k^2)u^3$, so that last term is $-\tfrac23 mg\,u^2/\sqrt{1-u^2}$ — the curvature term subtracts, as it must for a surface that curves away from the mass — and the formula follows. Checks: $u = 0$ gives $N = mg$, the mass resting on a horizontal tangent plane; $u \to 1$ (the vertical rim) gives $N \to -\infty$, meaning the surface would have to pull, which is why the mass has left long before.

01 · the claim · example

Three laws, three right-hand sides

worked example 1 — reduction to first order

equation of motion	state $y$	right-hand side $f(y,t)$	domain $\Omega$
free fall, $m\ddot z = -mg$	$(z, w)$	$(w,\ -g)$	$\mathbb{R}^2 \times \mathbb{R}$
pendulum, $\ddot\theta = -\tfrac{g}{\ell}\sin\theta$	$(\theta, u)$	$\big(u,\ -\tfrac{g}{\ell}\sin\theta\big)$	$\mathbb{R}^2 \times \mathbb{R}$
the dome, $\ddot s = k\operatorname{sgn}(s)\sqrt{\|s\|}$	$(s, v)$	$\big(v,\ k\operatorname{sgn}(s)\sqrt{\|s\|}\,\big)$	$\{\|s\| < g^2/k^2\} \times \mathbb{R}$

Step 1. Name the configuration variable and its velocity; that pair is $y$. Step 2. Write $\dot y_1 = y_2$ in the first block — always, for every mechanical system. Step 3. Put $F/m$, expressed in $y$ and $t$, in the second block. Step 4. Say where $f$ is defined: the dome's meridian is a signed arc length $s$, positive one way and negative the other, and the surface ends at $|s| = g^2/k^2$.

All three right-hand sides are continuous on their domains. Only the first two are Lipschitz, and that is the entire story of this unit.

Why the dome's equation acquires a sign. On a single meridian the mass can slide either way, so the natural coordinate is a signed arc length $s$, with $h(s) = -\tfrac{2k}{3g}|s|^{3/2}$. Then $h'(s) = -\tfrac{k}{g}\operatorname{sgn}(s)\sqrt{|s|}$ and $\ddot s = -g h'(s)$ is the entry in the table. For $s > 0$ it is $\ddot s = k\sqrt s$, which is the form we work with.

The dome is really two-dimensional. The mass may leave the apex down any meridian, so the true configuration space is the surface, and the family of section 3 gets an extra angular parameter: a circle's worth of directions times a half-line's worth of departure times. Restricting to one meridian loses none of the mathematics and keeps the picture in two dimensions of phase space.

Free fall is the degenerate case worth keeping in view. Its $f(z,w) = (w,-g)$ depends on the state only through the velocity, and only in the block that is a tautology: $\|f(y)-f(z)\| = |\Delta w| \le \|y-z\|$, with equality, so $L = 1$ exactly. Picard's iteration terminates after two steps, at $\big(z_0 + w_0 t - \tfrac12 g t^2,\ w_0 - gt\big)$, which is the exact answer — the second block is right after one step and the first after two, because each application of $P$ moves information up one block. Every theorem in this unit should be checked against it first.

01 · the claim

Three ways the future can fail to be fixed

1 · no solution

Some continuous-looking laws admit no solution at all through a given state. Rare in mechanics, and ruled out by mere continuity of $f$ (Peano, section 3).

2 · many solutions

One state, several futures. This is the real failure of determinism, and it needs the Lipschitz hypothesis to be broken. Section 3: the dome.

3 · the solution ends

A unique solution that ceases to exist after a finite time, having gone to infinity. Compatible with every hypothesis of the theorem. Section 4.

The theorem in section 2 closes the first two, locally and under a hypothesis. Nothing closes the third: it is a genuine feature of perfectly good laws, and the best available statement is a description of how a solution ends.

Where the unit is going. §2 the theorem · §3 what happens when its hypothesis fails · §4 how far its guarantee reaches, and how two futures drift apart inside it.

The three sketches. (1) The field $\dot y = +1$ below the axis and $-1$ above it: a solution rising to $y = 0$ cannot continue, since just above the axis it would have to fall and just below it would have to rise. (2) The dome's family: one state at $t = 0$, curves leaving it at every later time, in both directions. (3) $\dot x = x^2$ from $x(0) = 2$, which reaches infinity at $T_+ = 0.5$; it is plotted properly in section 4.

A fourth failure, not on the list because it is not a failure of the mathematics: the solution exists and is unique but is useless, because the map from initial data to solution, while single-valued, amplifies errors so fast that no measurement fixes the answer. That is unit 17. Section 4's continuous dependence estimate is exactly the bound that chaos saturates.

Failure 1 does happen, and the standard example is not a mechanical one: $\dot y = g(y)$ with $g$ discontinuous, say $g(y) = 1$ for $y \le 0$ and $g(y) = -1$ for $y > 0$, has no solution through $y(0) = 0$ in the classical sense — the field points inwards from both sides. Mechanics avoids this because forces are continuous, which is why Peano's theorem is enough to dispose of failure 1 in everything we shall meet.

section 2

Existence, and exactly one of them

A solution is a fixed point of an integral operator, and the Lipschitz condition makes that operator a contraction.

01 · the theorem

Step one: turn the ODE into an integral equation

lemma

Let $f$ be continuous on an open $\Omega \subseteq \mathbb{R}^N \times \mathbb{R}$, let $J$ be an interval containing $t_0$, and let $y : J \to \mathbb{R}^N$ be continuous with $(y(t), t) \in \Omega$. Then

$$y \in C^1(J) \ \text{ and }\ \dot y = f(y, t),\ y(t_0) = y_0 \qquad\Longleftrightarrow\qquad y(t) = y_0 + \int_{t_0}^{t} f\big(y(s), s\big)\,\mathrm{d}s \ \text{ on } J.$$

proof

($\Rightarrow$) $s \mapsto f(y(s),s)$ is continuous, being a composition of continuous maps, so the fundamental theorem of calculus applied componentwise gives $y(t) - y(t_0) = \int_{t_0}^t \dot y = \int_{t_0}^t f(y,s)\,\mathrm{d}s$.

($\Leftarrow$) The integrand is continuous, so the right-hand side is differentiable with derivative $f(y(t),t)$; hence $y$ is $C^1$ and solves the equation, and setting $t = t_0$ gives $y(t_0) = y_0$. $\square$

Why bother. Three gains, all decisive:

the initial condition is inside the equation, not a side condition;
the unknown appears under an integral, which smooths, where a derivative roughens;
the requirement on the candidate drops from $C^1$ to $C^0$, and $C^0$ with the sup norm is a complete metric space.

The whole proof is: the right-hand side is a map $P$ on continuous functions, and a solution is a point with $P y = y$.

Integrals of vector-valued functions are taken componentwise, and the one inequality we need is $\big\|\int_a^b \varphi\big\| \le \int_a^b \|\varphi\|$ for $a \le b$. Proof: put $w = \int_a^b \varphi$. Then $\|w\|^2 = \langle w, \int_a^b \varphi\rangle = \int_a^b \langle w, \varphi(s)\rangle\,\mathrm{d}s \le \int_a^b \|w\|\,\|\varphi(s)\|\,\mathrm{d}s$ by Cauchy–Schwarz, and dividing by $\|w\|$ (or noting the claim is trivial when $w = 0$) gives it. Linearity of the integral in the first slot is what lets $w$ come out.

The smoothing remark, concretely. If $y$ is merely continuous, $Py$ is $C^1$; if $y$ is $C^1$, $Py$ is $C^2$. So a fixed point is automatically as smooth as $f$ allows — the solution of $\ddot r = k\sqrt r$ turns out to be $C^3$ and no better, because $\sqrt{\cdot}$ is only so smooth, and that is visible from this remark alone.

The check. Both sides have the units of $y$, since $[f]\,[t] = [y]$. Special case $f = f(t)$: the equivalence is the fundamental theorem of calculus itself, and the integral equation is already the answer — unit 00's first quadrature case, with nothing left to solve.

This reformulation is Picard's (Émile Picard, 1890, méthode des approximations successives). Lindelöf's 1894 paper sharpened the hypotheses to the form used today; in French the result is still called the Cauchy–Lipschitz theorem, after the two people who had it first, in 1824 and 1868 respectively, by a different argument.

01 · the theorem

The Lipschitz condition

Lipschitz in $y$

$f$ is Lipschitz in $y$ with constant $L$ on $S \subseteq \mathbb{R}^N \times \mathbb{R}$ if

$$\|f(y,t) - f(z,t)\| \le L\,\|y - z\|$$

for every pair $(y,t), (z,t) \in S$ — both taken at the same $t$.

$f$ is locally Lipschitz in $y$ on an open $\Omega$ if every point of $\Omega$ has a neighbourhood on which such an $L$ exists. The constant may differ from place to place; it may not be infinite anywhere.

Same $t$ on both sides: no regularity in $t$ beyond continuity is asked for, and none is used.

Lipschitz with constant $L$: wherever the vertex of the wedge of slopes $\pm L$ is put on the graph, the graph stays inside it. Here $\sin\theta$ and $L = 1$.

The picture behind the definition. For a function of one variable, $|g(b) - g(a)| \le L|b - a|$ for all $b$ says that the graph lies inside the double wedge of slopes $\pm L$ with its vertex at $(a, g(a))$. Lipschitz with constant $L$ means: put the vertex anywhere on the graph, and the whole graph stays inside. For $\sin$ the wedge of slopes $\pm 1$ works at every vertex, which is the mean value bound on the next slide. For a vector-valued $f$ the wedge becomes a cone, $\|f(y) - f(z)\| \le L\|y - z\|$, and the picture is the same.

Why "Lipschitz in $y$, uniformly in $t$" and not something weaker. The estimate we shall make integrates $\|f(y(s),s) - f(z(s),s)\|$ over $s$, comparing the two curves at the same instant. Nothing is ever compared at different times, so no continuity in $t$ beyond what makes the integrand integrable is needed. This is why the theorem applies unchanged to a force switched on abruptly, as long as it is continuous.

01 · the theorem

Lipschitz constants, checked by hand

checked by hand

Pendulum, dimensionless form $f(\theta,u) = (u, -\sin\theta)$. Since $|\sin a - \sin b| \le |a - b|$ by the mean value theorem, $\|f(y)-f(z)\|^2 = (\Delta u)^2 + (\sin\theta_y - \sin\theta_z)^2 \le (\Delta u)^2 + (\Delta\theta)^2$, so $L = 1$ on all of $\mathbb{R}^2$.

Square root, $g(s) = \sqrt s$ on $[0,\infty)$: $\dfrac{|g(s)-g(0)|}{|s-0|} = \dfrac{1}{\sqrt s} \to \infty$ as $s \to 0^+$. No $L$ works on any neighbourhood of $0$; on $[\epsilon,\infty)$, $L = 1/(2\sqrt\epsilon)$ does.

not available yet

"$f$ is $C^1$, hence locally Lipschitz" needs the mean value inequality in several variables, which arrives in unit 03. Until then every Lipschitz constant in this course is produced by hand.

The square root at $0$: every line $Ls$ through the origin is beaten by $\sqrt s$ on $0 < s < 1/L^2$ — red for $L = 2$. Raising $L$ shrinks the interval and never removes it.

The pendulum bound in full. $|\sin a - \sin b| = |\cos \xi|\,|a-b| \le |a-b|$ for some $\xi$ between $a$ and $b$, by the single-variable mean value theorem — which the learner has. In dimensional form $f(\theta, u) = (u, -(g/\ell)\sin\theta)$ and $L = \max(1, g/\ell)$ works, but the two entries of that maximum have different dimensions, which is the warning from the phase-space slide arriving on schedule. Nondimensionalizing with $\tau = \sqrt{\ell/g}$ makes $L = 1$, a pure number.

Reading the figure. A line $Ls$ through the origin lies below $\sqrt s$ exactly when $\sqrt s > Ls$, i.e. $s < 1/L^2$. So every candidate constant fails on an interval next to $0$ — a shorter interval for a larger $L$, but never an empty one. That is "no $L$ works on any neighbourhood of $0$", drawn.

The square root is Hölder, not Lipschitz. $|\sqrt a - \sqrt b| \le \sqrt{|a-b|}$ for $a, b \ge 0$: proof, assume $a \ge b$, then $(\sqrt a - \sqrt b)^2 \le (\sqrt a - \sqrt b)(\sqrt a + \sqrt b) = a - b$. So $\sqrt{\cdot}$ is Hölder continuous with exponent $\tfrac12$. Section 3 shows that exponent is on the wrong side of the line.

01 · the theorem

The contraction mapping principle

Banach fixed-point theorem

Let $(X, d)$ be a non-empty complete metric space and $P : X \to X$ a contraction: $d(Pu, Pw) \le q\, d(u,w)$ for all $u,w$, with a fixed $q < 1$. Then $P$ has exactly one fixed point $u^\ast$, and for every $u_0 \in X$ the iterates $u_{n+1} = P u_n$ converge to it, with

$$d(u_n, u^\ast) \le \frac{q^n}{1-q}\, d(u_1, u_0).$$

The theorem is constructive: it does not merely assert a solution, it names a sequence converging to it and bounds the error at every step. That is the Picard iteration, and worked example 2 runs it.

Iterating $P(u) = \cos u$ on $[0,1]$, a contraction with $q = \sin 1 = 0.84$: up to the graph, across to the diagonal, repeat. From $u_0 = 0$ the path closes in on the one fixed point, $u^\ast = 0.739$.

01 · the theorem

The contraction principle: proof, and how to lose it

proof

$d(u_{n+1},u_n) \le q^n d(u_1,u_0)$ by induction, so for $m > n$, $d(u_m,u_n) \le \sum_{j=n}^{m-1} q^j d(u_1,u_0) \le \frac{q^n}{1-q} d(u_1,u_0)$: the sequence is Cauchy, hence convergent to some $u^\ast$. $P$ is continuous (it is $q$-Lipschitz), so $Pu^\ast = \lim P u_n = \lim u_{n+1} = u^\ast$. If $Pu^\ast = u^\ast$ and $Pw = w$ then $d(u^\ast,w) \le q\,d(u^\ast,w)$, forcing $d = 0$. Letting $m \to \infty$ in the displayed bound gives the estimate. $\square$

both hypotheses earn their place

Completeness: on $X = (0,1]$ with $Pu = u/2$, $q = 1/2$ and there is no fixed point.

$q < 1$, not $d(Pu,Pw) < d(u,w)$: on $X = [1,\infty)$, $Pu = u + 1/u$ strictly decreases distances and has no fixed point.

$Pu = u + 1/u$ on $[1,\infty)$ stays above the diagonal by $1/u$, a gap that shrinks and never closes: no fixed point.

01 · the theorem

The space the contraction acts on

Fix $J = [t_0 - a, t_0 + a]$ and let $C(J, \mathbb{R}^N)$ carry the sup norm $\|y\|_\infty = \sup_{t \in J}\|y(t)\|$.

completeness

$\big(C(J,\mathbb{R}^N), \|\cdot\|_\infty\big)$ is a complete metric space, and so is every closed subset of it — in particular the tube

$$X = \big\{\, y \in C(J,\mathbb{R}^N) : \|y - y_0\|_\infty \le b \,\big\}.$$

$X$ is non-empty: the constant function $y \equiv y_0$ lies in it.

weighted sup norm

For $\lambda > 0$, put $\ \|y\|_\lambda = \sup_{t \in J} e^{-\lambda|t - t_0|}\,\|y(t)\|$.

On the compact $J$, $\;e^{-\lambda a}\|y\|_\infty \le \|y\|_\lambda \le \|y\|_\infty$, so the two norms are equivalent: same open sets, same Cauchy sequences, same complete subsets. $X$ is complete in $\|\cdot\|_\lambda$ too.

$\lambda$ discounts the future. It will be chosen large enough to make the operator a contraction no matter how large $L$ is — the trick that keeps the existence interval as long as the geometry allows.

Full proof of completeness

Let $(y_n)$ be Cauchy in $\|\cdot\|_\infty$. For each fixed $t$, $\|y_n(t) - y_m(t)\| \le \|y_n - y_m\|_\infty$, so $(y_n(t))$ is Cauchy in $\mathbb{R}^N$, which is complete; call the limit $y(t)$. Given $\varepsilon > 0$ choose $n_0$ with $\|y_n - y_m\|_\infty \le \varepsilon$ for $n,m \ge n_0$; letting $m \to \infty$ in $\|y_n(t)-y_m(t)\| \le \varepsilon$ gives $\|y_n(t) - y(t)\| \le \varepsilon$ for every $t$, i.e. $y_n \to y$ uniformly. A uniform limit of continuous functions is continuous: given $t$ and $\varepsilon$, pick $n$ with $\|y_n - y\|_\infty \le \varepsilon/3$ and then $\delta$ with $\|y_n(t') - y_n(t)\| \le \varepsilon/3$ for $|t'-t| \le \delta$; the triangle inequality does the rest. So $y \in C(J,\mathbb{R}^N)$ and $\|y_n - y\|_\infty \to 0$.

$X$ is closed: if $y_n \to y$ uniformly and $\|y_n(t) - y_0\| \le b$ for all $n, t$, then $\|y(t) - y_0\| \le b$ for all $t$ by taking limits. A closed subset of a complete space is complete, because a Cauchy sequence in it converges in the big space and its limit lies in the subset.

The norm equivalence: $e^{-\lambda|t-t_0|} \in [e^{-\lambda a}, 1]$ for $t \in J$, and both inequalities follow by taking suprema. Equivalent norms have the same Cauchy sequences, so completeness transfers — this is the only property of $\|\cdot\|_\lambda$ we need besides the estimate on the next slide.

The weighted norm is due to Bielecki (1956), which is why it is often called the Bielecki norm. It is a later simplification: Picard and Lindelöf shrank the interval until $aL < 1$ instead.

01 · the theorem

Picard–Lindelöf

theorem · Picard–Lindelöf

Let $\Omega \subseteq \mathbb{R}^N \times \mathbb{R}$ be open, $f : \Omega \to \mathbb{R}^N$ continuous and locally Lipschitz in $y$, and $(y_0,t_0) \in \Omega$. Choose $b, c > 0$ so that the compact cylinder $\;\mathcal{C} = \{(y,t) : \|y - y_0\| \le b,\ |t - t_0| \le c\}\;$ lies in $\Omega$, and set

$$M = \max_{\mathcal{C}} \|f\|, \qquad L = \text{a Lipschitz constant for } f \text{ in } y \text{ on } \mathcal{C}, \qquad a = \min\Big(c, \frac{b}{M}\Big).$$

Then the initial value problem $\dot y = f(y,t)$, $y(t_0) = y_0$ has exactly one solution on $J = [t_0 - a,\ t_0 + a]$.

Where $a$ comes from. A solution moves at speed at most $M$, so it stays in the double cone of slopes $\pm M$. It is safe until the cone leaves the cylinder: through the sides at $|t - t_0| = c$, or through the top and bottom at $|t - t_0| = b/M$, whichever comes first.

No $L$ in the length of $J$: that is pure geometry.

The cylinder $\mathcal{C}$ and the cone of slopes $\pm M$ a solution cannot leave. $J$ is where the cone is still inside $\mathcal{C}$.

Why $a = \min(c, b/M)$ and not something larger. Two things can go wrong as $t$ moves away from $t_0$: the time can leave the cylinder ($|t - t_0| > c$), or the state can ($\|y(t)-y_0\| > b$). The state moves at speed at most $M$, so it needs at least $b/M$ to escape. Taking the minimum guarantees neither happens on $J$. This is the only place the length of the interval is decided, and it is pure geometry: no $L$ appears.

Reading the figure. It is drawn for $N = 1$, with $y$ vertical and $t$ horizontal. The shaded box is the cylinder $\mathcal{C}$; the darker double wedge is $\{|y - y_0| \le M|t - t_0|\}$, where any solution must stay because $\|\dot y\| = \|f\| \le M$ inside $\mathcal{C}$. On the left $M$ is small and the wedge leaves through the sides, so $a = c$; on the right $M$ is large and the wedge leaves through the top and bottom first, so $a = b/M$ and $J$ (the thick bar) is shorter than the cylinder.

01 · the theorem

Picard–Lindelöf: checks, and what it does not say

checks

Dimensions: $[M] = [y]/[t]$, so $b/M$ is a time, as $a$ must be.

Special case $f = f(t)$: $L = 0$, and the unique solution is $y_0 + \int_{t_0}^t f$ — unit 00's quadrature, recovered.

Limit $M \to \infty$ (a violent force): $a \to 0$. The guarantee is short exactly when the state leaves the cylinder fast.

$M$ and $L$ are finite because $\mathcal{C}$ is compact, $f$ is continuous, and a locally Lipschitz function has a single Lipschitz constant on any compact subset of its domain.

Read the conclusion carefully. It is local: a guarantee on $J$, and nothing beyond. Section 4 says how far $J$ can be pushed and what stops it.

proof over the next two slides; the compactness lemma is in the notes (N)

Full proof that a locally Lipschitz

Full proof that a locally Lipschitz $f$ has one constant on a compact set

Suppose not: then there are $(y_n,t_n), (z_n,t_n) \in \mathcal{C}$ with $\|f(y_n,t_n)-f(z_n,t_n)\| > n\|y_n - z_n\|$. Since $f$ is bounded on $\mathcal{C}$ by $M$, the left side is at most $2M$, so $\|y_n - z_n\| < 2M/n \to 0$. By compactness pass to a subsequence with $(y_n,t_n) \to (y_\ast,t_\ast) \in \mathcal{C}$, and then $(z_n,t_n) \to (y_\ast,t_\ast)$ as well. Take a neighbourhood $U$ of $(y_\ast,t_\ast)$ on which $f$ is Lipschitz in $y$ with some constant $L_\ast$; for large $n$ both points lie in $U$, giving $n\|y_n-z_n\| < L_\ast\|y_n - z_n\|$, absurd once $n > L_\ast$ unless $y_n = z_n$ — and if $y_n = z_n$ the strict inequality fails outright. $\square$

Hypotheses that are not there. No linearity, no autonomy, no smoothness beyond continuity in $t$, no bound on $f$ at infinity, and no conservation law. That is why the theorem covers every force law in this course, including the ones with no formula for their solutions.

01 · the theorem · proof

Proof, part 1: the operator maps the tube into itself

On $X = \{y \in C(J,\mathbb{R}^N) : \|y(t) - y_0\| \le b\}$ define the Picard operator

$$(Py)(t) = y_0 + \int_{t_0}^{t} f\big(y(s), s\big)\,\mathrm{d}s .$$

$P$ is well defined and $P(X) \subseteq X$

For $y \in X$ and $s \in J$, the point $(y(s),s)$ lies in $\mathcal{C} \subseteq \Omega$, so the integrand is defined, and it is continuous in $s$; hence $Py \in C(J,\mathbb{R}^N)$. Moreover

$$\|(Py)(t) - y_0\| \le \Big|\int_{t_0}^{t}\|f\|\,\mathrm{d}s\Big| \le M\,a \le b,$$

using $|t-t_0| \le a$ and then $a \le b/M$. So $Py \in X$. $\square$

Computed for $f(y,t) = \cos(t + 2y)$, $M = 1$, $b = a = 1$. A wiggly candidate $y$ in the tube; its image $Py$ starts at $y_0$, has slope at most $M$, so stays in the dashed cone — hence in the tube. Note how much smoother $Py$ is.

what has been bought

Solutions in the tube are exactly the fixed points of $P$ (the lemma), and $P$ maps the complete $X$ to itself. Banach needs one thing more, a contraction factor: the next slide, the only use of Lipschitz.

The figure, computed. Scalar case, $t_0 = 0$, $y_0 = 0$, $f(y,t) = \cos(t + 2y)$, so $|f| \le 1 = M$ everywhere and the cylinder with $b = 1$, $c = 1.5$ gives $a = \min(1.5, 1/1) = 1$. The candidate is $y(t) = 0.8\sin(4.5t + 0.6)$, which is in $X$ although it does not even start at $y_0$; $Py$ was found by numerical quadrature of $\int_0^t \cos(s + 2y(s))\,\mathrm{d}s$. Two things to see: $Py$ always starts at $y_0$, whatever $y$ was, and $Py$ has bounded slope, so it is far tamer than $y$ — the smoothing of the integral, visible.

Where $M$ went. Into making $P$ land in $X$, and nowhere else: it plays no part in the contraction on the next slide, just as $L$ plays no part here. $X$ is non-empty because the constant $y \equiv y_0$ lies in it.

The absolute value around the integral is not decoration. For $t < t_0$ the integral runs backwards and $\int_{t_0}^t = -\int_t^{t_0}$; the bound $\big\|\int_{t_0}^t \varphi\big\| \le \big|\int_{t_0}^t\|\varphi\|\big|$ holds in both directions, and every estimate in this proof is symmetric in the two time directions. Determinism runs backwards as firmly as forwards — Laplace said so, and this is where it comes from.

Continuity of $s \mapsto f(y(s),s)$: it is the composition of the continuous $s \mapsto (y(s),s)$ with the continuous $f$. Nothing deeper. And it is bounded on the compact $J$, so the integral exists as a Riemann integral, componentwise.

Why the tube and not all of $C(J,\mathbb{R}^N)$. Because $f$ need not be defined, let alone Lipschitz, outside $\mathcal{C}$. The tube is the largest set of candidate curves on which the operator makes sense, and $a \le b/M$ is precisely the condition that $P$ does not push a candidate out of it.

01 · the theorem · proof

Proof, part 2: the weighted norm makes it a contraction

the estimate

Take $y, z \in X$ and $t \ge t_0$ (the case $t \le t_0$ is identical with the limits reversed). Using the Lipschitz condition on $\mathcal{C}$ and then inserting $1 = e^{\lambda(s-t_0)}e^{-\lambda(s-t_0)}$:

$$\|(Py)(t)-(Pz)(t)\| \le \int_{t_0}^{t}\! L\|y(s)-z(s)\|\,\mathrm{d}s = \int_{t_0}^{t}\! L\,e^{\lambda(s-t_0)}\Big(e^{-\lambda(s-t_0)}\|y(s)-z(s)\|\Big)\mathrm{d}s$$ $$\le L\,\|y-z\|_\lambda \int_{t_0}^{t}\! e^{\lambda(s-t_0)}\,\mathrm{d}s = \frac{L}{\lambda}\Big(e^{\lambda(t-t_0)} - 1\Big)\|y-z\|_\lambda \;\le\; \frac{L}{\lambda}\,e^{\lambda(t-t_0)}\|y-z\|_\lambda .$$

Multiply by $e^{-\lambda(t-t_0)}$ and take the supremum over $t \in J$: $\;\|Py - Pz\|_\lambda \le \dfrac{L}{\lambda}\|y - z\|_\lambda$. Choosing $\lambda = 2L$ gives $q = \tfrac12 < 1$. $\square$

conclusion

$X$ complete, $P : X \to X$ a contraction $\Rightarrow$ exactly one fixed point $\Rightarrow$ exactly one solution in the tube on $J$. With the lemma, Picard–Lindelöf is proved.

what the weight rescued

Without it the estimate reads $\|Py-Pz\|_\infty \le La\,\|y-z\|_\infty$, a contraction only if $a < 1/L$ — an interval that shrinks as the force gets stiffer, for no good reason. The weight makes $L$ irrelevant to the length of the interval, which stays $\min(c, b/M)$.

The step that is easy to miss: $e^{-\lambda(s-t_0)}\|y(s)-z(s)\| \le \|y-z\|_\lambda$ holds for every $s \in J$ by the definition of the weighted norm as a supremum. That is the whole trick: the exponential weight pays for the growth of the integral, and $\lambda$ is ours to choose.

The case $t \le t_0$. Write $\big\|\int_{t_0}^t \varphi\big\| \le \int_t^{t_0}\|\varphi\|$ and repeat with $e^{\lambda(t_0-s)}$; the weight $e^{-\lambda|t-t_0|}$ is symmetric, so the same factor $L/\lambda$ comes out. Doing both halves at once with $|t - t_0|$ everywhere is possible and less readable.

Uniqueness so far is uniqueness in $X$. A solution on $J$ that wanders out of the tube is not yet excluded by this argument. It is excluded on the next slide, by a connectedness argument that also gives uniqueness on intervals far longer than $J$.

Error bound, for free. Banach's estimate with $q = 1/2$ gives $\|y_n - y\|_\lambda \le 2^{1-n}\|y_1-y_0\|_\lambda$, and $\|y_1 - y_0\|_\lambda \le \|y_1-y_0\|_\infty \le Ma$. So the $n$-th Picard iterate is within $2^{1-n}Ma\,e^{\lambda a}$ of the solution in sup norm. Worked example 2 compares this with the truth in a case where the truth is known.

01 · the theorem

Uniqueness, beyond the interval the proof gave

global uniqueness

Let $f$ be continuous and locally Lipschitz in $y$ on an open $\Omega$, and let $y, z$ solve $\dot w = f(w,t)$ on a common interval $I$ with $y(t_1) = z(t_1)$ for some $t_1 \in I$. Then $y = z$ on all of $I$.

proof

Let $A = \{t \in I : y(t) = z(t)\}$. It is non-empty ($t_1 \in A$) and closed in $I$, being the preimage of $\{0\}$ under the continuous $y - z$. It is open in $I$: at $t_\ast \in A$ apply the theorem at $(y(t_\ast), t_\ast)$; both $y$ and $z$ are continuous, so both stay in the tube for a short time either side, and there they are the same fixed point. $I$ is connected, so $A = I$. $\square$

So there is no such thing as two solutions that agree for a while and then part. Solutions through a common point coincide on every interval where both live.

what this licenses

Guess a formula, check it satisfies the equation and the initial data, and you have found the solution — no further argument needed. Unit 06 will guess exponentials; unit 08 will guess conics.

$x(t) = \cos \omega t$ solves $\ddot x = -\omega^2 x$, $x(0)=1$, $\dot x(0)=0$, as substitution shows, and it is now the only one.

The open step, in detail. Let $t_\ast \in A$, $w_\ast = y(t_\ast) = z(t_\ast)$. Build the cylinder $\mathcal{C}$ around $(w_\ast,t_\ast)$ and let $a$ be as in the theorem. By continuity of $y$ and of $z$ there is $\delta \in (0,a]$ with $\|y(t)-w_\ast\| \le b$ and $\|z(t)-w_\ast\| \le b$ for $|t-t_\ast| \le \delta$. On that interval both are fixed points of the Picard operator built at $(w_\ast, t_\ast)$ — the operator on $[t_\ast-\delta,t_\ast+\delta]$, whose contraction constant is still $L/\lambda$ — so they are equal. Hence $A$ contains a relative neighbourhood of $t_\ast$.

Connectedness is doing real work. On a disconnected domain the conclusion is false for the silly reason that the two pieces know nothing of each other. This is also why maximal solutions, in section 4, live on intervals.

The licence is the point of the unit. Almost every explicit solution in the rest of this course is found by guessing an ansatz and verifying it. Without uniqueness, verification shows only that a formula is a solution, and a physicist who says "the trajectory" is making an unsupported claim. After this slide, the claim is supported — wherever the Lipschitz condition holds.

01 · the theorem · example

Picard's iteration, run

worked example 2 — $\ddot x = -x$, $x(0)=1$, $\dot x(0)=0$

Step 1. Phase space: $y = (x,v)$, $f(y) = (v, -x)$, $y_0 = (1,0)$. Lipschitz with $L=1$ on $\mathbb{R}^2$, as on the pendulum slide.

Step 2. Start from the constant $y^{(0)}(t) \equiv (1,0)$ and iterate $y^{(n+1)}(t) = (1,0) + \int_0^t f\big(y^{(n)}(s)\big)\,\mathrm{d}s$:

$y^{(1)} = (1,0) + \int_0^t (0,-1)\,\mathrm{d}s = (1,\,-t)$;

$y^{(2)} = (1,0) + \int_0^t (-s,\,-1)\,\mathrm{d}s = \big(1 - \tfrac{t^2}{2},\ -t\big)$;

$y^{(3)} = (1,0) + \int_0^t \big(-s,\ -1+\tfrac{s^2}{2}\big)\mathrm{d}s = \big(1 - \tfrac{t^2}{2},\ -t + \tfrac{t^3}{6}\big)$.

Step 3. The iterates are the Taylor partial sums of $(\cos t, -\sin t)$ — visibly, and by induction on the two blocks.

Step 4. At $t = 1$: $x^{(2)} = 0.5$ against $\cos 1 = 0.540302$, error $0.040302$, inside the alternating-series bound $1/4! = 0.041667$. At $y^{(4)}$: $x^{(4)} = 0.541667$, error $0.001364$, bound $1/6! = 0.001389$.

$\cos t$ and the first block $x^{(n)}$ of the iterates: each follows the solution from $t = 0$ and peels away later the larger $n$ is.

Reading the figure. $x^{(n)}$ is the first block of the $n$-th iterate, the Taylor polynomial of $\cos$ of degree $n$, plotted for $n = 2, 4, 6, 8$. Near $t = 0$ all of them lie on $\cos t$; each leaves it once its first omitted term, $t^{n+2}/(n+2)!$, stops being small, which happens later the larger $n$ is. At $t = 1$ (the dotted line) the errors are the ones in Step 4.

The induction. Claim: $y^{(2m)} = \big(\sum_{j=0}^{m}\frac{(-1)^j t^{2j}}{(2j)!},\ \sum_{j=0}^{m-1}\frac{(-1)^{j+1}t^{2j+1}}{(2j+1)!}\big)$, and $y^{(2m+1)}$ has the same first block and one more term in the second. Each application of $P$ integrates the second block into the first and minus the first into the second, which shifts the partial sums by one term. The limit is $(\cos t, -\sin t)$, whose series converge everywhere.

The alternating-series bound applies because $\cos 1 = \sum (-1)^j/(2j)!$ has terms decreasing in magnitude to zero, so truncating after the $t^2$ term leaves an error at most the next term, $1/4!$. The two numerical checks above were computed, not quoted.

01 · the theorem · example

What the iteration shows

what the example shows

The theorem's proof is an algorithm. Picard iteration converges here on all of $\mathbb{R}$, faster than the geometric rate Banach guarantees, because $f$ is linear.

Dimension check: in dimensionless time. With $\ddot x = -\omega^2 x$ the iterates are partial sums in the dimensionless variable $\omega t$, as they must be.

Limit check: set the force to zero and the iteration stops at $y^{(1)} = (1, 0)$, the constant solution — correct.

Uniqueness (previous slide) is what turns "$\cos$ is the limit of the iterates" into "$\cos$ is the motion".

The iterates in phase space, $0 \le t \le 2\pi$: the solution is the circle from $y_0 = (1,0)$; $n = 2$ (dotted) leaves it at once, $n = 8$ (solid) only after most of a turn.

Reading the figure. The same iterates, now as curves $t \mapsto y^{(n)}(t) = (x^{(n)}, v^{(n)})$ in the phase plane, for $0 \le t \le 2\pi$. The solution is the unit circle, traced clockwise from $y_0 = (1, 0)$. $y^{(2)} = (1 - t^2/2,\ -t)$ is a parabola that leaves the circle almost at once; each later iterate hugs the circle for longer before spinning off. On any bounded time interval the iterates converge uniformly — that is the content of the theorem — and the picture shows why no single $n$ works for all time.

Why the convergence beats the guarantee. Banach promises $2^{1-n}$ on an interval of length $a$; the true error is $O(t^{n}/n!)$. The reason is that the iteration's error satisfies $\|y^{(n+1)}-y\|(t) \le \int_0^t L\|y^{(n)}-y\|$, which iterated gives $L^n t^n/n!$, not $(Lt)^n$. One can build this sharper factorial estimate into the theorem instead of using the weighted norm; it is the traditional route, and it gives the same interval.

01 · the theorem · the dome

The dome passes the test, except at one point

The dome's right-hand side is $f(s,v) = \big(v,\ k\operatorname{sgn}(s)\sqrt{|s|}\big)$.

away from the apex

On $S_\epsilon = \{(s,v) : \epsilon \le |s| \le g^2/k^2\}$, the mean value theorem gives $\big|\sqrt{a}-\sqrt{b}\big| \le \frac{1}{2\sqrt\epsilon}|a-b|$ for $a,b \ge \epsilon$, and $\sqrt x \le x/\sqrt\epsilon$ handles a pair straddling the apex, so $f$ is Lipschitz there with $L = \max\big(1, \tfrac{k}{\sqrt\epsilon}\big)$.

Picard–Lindelöf applies at every state with $s \ne 0$: a mass released anywhere on the slope has exactly one future and exactly one past.

Left, the force per unit mass, $k = 1$: continuous, with a vertical tangent at $s = 0$. Right, its slope $k|s|^{-1/2}/2$: bounded by $k\epsilon^{-1/2}$ once the strip $|s| < \epsilon$ is cut out, unbounded at the apex.

at the apex

$\dfrac{\|f(s,0)-f(0,0)\|}{\|(s,0)-(0,0)\|} = \dfrac{k\sqrt{s}}{s} = \dfrac{k}{\sqrt s} \longrightarrow \infty$ as $s \to 0^+$.

At $s = 10^{-6}\,\mathrm{m}$ the ratio is already $10^{3}\,\mathrm{s^{-2}}$, and no neighbourhood of the apex admits any finite $L$. The hypothesis fails at exactly one point of phase space — the one we care about.

Reading the figure. The left panel is the second block of $f$ against $s$ (it does not depend on $v$). The right panel is its derivative, $\frac{\mathrm{d}}{\mathrm{d}s}\,k\operatorname{sgn}(s)\sqrt{|s|} = k/(2\sqrt{|s|})$ for $s \ne 0$: the local steepness of the force. Lipschitz on a set means, roughly, steepness bounded on that set. Remove the shaded strip and it is bounded; keep any piece of the apex and it is not. $f$ is still continuous there, and $\operatorname{sgn}(s)\sqrt{|s|}$ is still Hölder-$\tfrac12$: it is the derivative of the square root, not the square root, that blows up.

The mean value estimate, written out. For $b \ge a \ge \epsilon > 0$, $\sqrt b - \sqrt a = \frac{b-a}{2\sqrt\xi}$ for some $\xi \in (a,b)$, and $\xi \ge \epsilon$, so the difference is at most $(b-a)/(2\sqrt\epsilon)$. The two branches. $S_\epsilon$ has a piece on each side of the apex, and a pair $s_1 \ge \epsilon \ge -s_2$ is not covered by that estimate: there $\big|\!\operatorname{sgn}(s_1)\sqrt{|s_1|} - \operatorname{sgn}(s_2)\sqrt{|s_2|}\big| = \sqrt{s_1}+\sqrt{|s_2|}$, and since $\sqrt x \le x/\sqrt\epsilon$ for $x \ge \epsilon$ this is at most $(|s_1|+|s_2|)/\sqrt\epsilon = |s_1-s_2|/\sqrt\epsilon$. The constant $k/\sqrt\epsilon$ covers both cases; combining the two blocks of $f$ as on the Lipschitz slide gives the stated $L$, and it diverges as $\epsilon \to 0$ — which is the content of the box on the right, said the other way round.

Read the two boxes together. The dome is not a wild object. Every state except one has a unique future; determinism holds on all of phase space minus a single point. That is what makes it disturbing: a single point of failure, placed exactly where a mass would naturally be put.

An honest alternative the dome does not take. One could declare the apex outside the domain, $\Omega = \{s \ne 0\}$, and then the theorem applies everywhere in $\Omega$ — but the physical surface is perfectly smooth-looking at the apex, the force there is exactly zero, and refusing to consider a mass sitting at the apex is refusing to consider the question.

section 3

The dome, taken apart

One state, a one-parameter family of futures, and exactly which hypothesis is missing.

01 · the dome · example

The one-parameter family, verified

worked example 3 — every member solves $\ddot r = k\sqrt r$ with the same data

The family. For each $T \ge 0$ define $\;r_T(t) = 0$ for $t \le T$, and $\;r_T(t) = \dfrac{k^2}{144}(t-T)^4$ for $t \ge T$. Also $r_\infty \equiv 0$.

Step 1 — it is $C^2$. For $t > T$: $\dot r_T = \frac{k^2}{36}(t-T)^3$ and $\ddot r_T = \frac{k^2}{12}(t-T)^2$. All three expressions tend to $0$ as $t \downarrow T$, and all three vanish identically for $t < T$, so $r_T$, $\dot r_T$, $\ddot r_T$ are continuous at $T$. (In fact $r_T \in C^3$: the third derivative is $\frac{k^2}{6}(t-T) \to 0$. The fourth jumps from $0$ to $k^2/6$.)

Step 2 — it solves the equation. For $t > T$, $\;k\sqrt{r_T} = k\cdot\frac{k}{12}(t-T)^2 = \frac{k^2}{12}(t-T)^2 = \ddot r_T$. ✓ For $t < T$ both sides are $0$. ✓ At $t = T$ both sides are $0$ and both are continuous. ✓

Step 3 — the data. $r_T(0) = 0$ and $\dot r_T(0) = 0$ for every $T \ge 0$: in phase space every member starts at the same point.

Step 4 — the energy. $E/m = \tfrac12\dot r^2 + g h(r) = \frac{k^4(t-T)^6}{2592} - \frac{k^4(t-T)^6}{2592} = 0$, the same value as for $r \equiv 0$. Nothing is created.

checks

Dimensions: $[k^2 t^4] = \mathrm{L\,T^{-4}\cdot T^4} = \mathrm{L}$. ✓

Limit $T \to \infty$: the family degenerates to the rest solution, which is therefore not a separate case but the endpoint of the family.

Number: with $k = 1\,\mathrm{m^{1/2}s^{-2}}$, the moving member reaches the rim $r = 96.17$ m at $t - T = (144 \times 96.17)^{1/4} = 10.85$ s.

Three members of the family, hovering shows them: same start, different departure times, identical shape thereafter.

Where the quartic comes from, rather than being pulled out of a hat. Look for a power solution $r = \alpha t^p$ of $\ddot r = k \sqrt r$. Then $\alpha p(p-1)t^{p-2} = k\sqrt\alpha\, t^{p/2}$, so $p - 2 = p/2$, giving $p = 4$; and $12\alpha = k\sqrt\alpha$ gives $\sqrt\alpha = k/12$, $\alpha = k^2/144$. Translating $t \to t - T$ is allowed because the equation has no explicit $t$ in it, and gluing to $0$ on the left is allowed because the glue point is $C^2$ — which is Step 1. This is unit 00's scaling habit used to find a solution rather than to check one.

The energy computation in full. $\tfrac12 \dot r^2 = \tfrac12\big(\frac{k^2}{36}\big)^2 (t-T)^6 = \frac{k^4 (t-T)^6}{2592}$. And $g h(r) = -\frac{2k}{3}r^{3/2} = -\frac{2k}{3}\big(\frac{k^2}{144}\big)^{3/2}(t-T)^{6} = -\frac{2k}{3}\cdot\frac{k^3}{1728}(t-T)^6 = -\frac{k^4(t-T)^6}{2592}$. They cancel exactly. So the spontaneous solution has the same energy, at every instant, as the mass that never moves: no energy conservation law is violated, and no one who objects on energy grounds has looked.

Reversibility. The equation is unchanged under $t \mapsto -t$, so the time reverse of the family is a mass launched up the slope with exactly the escape speed, arriving at the apex, at rest, in finite time, and staying. That motion surprises nobody. The dome's scandal is the same motion, watched backwards.

01 · the dome

The family, drawn

$r_T(t)$ for $T = 0, 2, 4, 6$ s and $r \equiv 0$. One state at $t = 0$, one shape, any delay.

Phase plane $(s, v)$, curves of constant energy. At $E = 0$ four branches meet at the apex (red): two leave it, two arrive.

Left: what the mass does. Right: where its state goes. In time the members differ; in phase space every one of them travels the same branch out of $(0,0)$ — and nothing about the point $(0,0)$ says when to start.

The left panel plots $r_T(t) = \frac{k^2}{144}(t-T)^4$ for $t \ge T$, zero before, with $k = 1\,\mathrm{m^{1/2}s^{-2}}$ and $T = 0, 2, 4, 6$ s, together with the rest solution $r \equiv 0$. Every curve is the same quartic shifted right; each joins the axis with zero slope and zero curvature, which is the $C^2$ glue of Step 1.

The right panel is the phase plane of the signed coordinate, $(s, v)$, with the curves of constant energy per unit mass $E = \tfrac12 v^2 - \tfrac{2k}{3}|s|^{3/2}$ (energy is conserved, unit 00). Negative $E$: the mass comes up the slope, stops short of the apex and slides back. Positive $E$: it passes over the top. $E = 0$ is four branches $v = \pm\sqrt{4k/3}\,|s|^{3/4}$ meeting at the apex; the two leaving it carry the family (one for each direction), and the two arriving carry its time reverse, a mass that reaches the apex, at rest, in finite time. A state on a leaving branch has exactly one future; only the apex itself, where all four meet, has many.

Why the branches meet the apex vertically. Along them $\mathrm{d}v/\mathrm{d}s \propto |s|^{-1/4} \to \infty$. Compare a Lipschitz equilibrium such as the inverted pendulum's, where the branches arrive along straight lines and take infinite time to get there.

01 · the dome

What exactly is indeterminate

the failure, stated

The initial value problem $\ddot s = k\operatorname{sgn}(s)\sqrt{|s|}$, $s(0)=0$, $\dot s(0)=0$ has uncountably many solutions on $[0,\infty)$: one for each departure time $T \in [0,\infty]$, and each may go either way along the meridian, and on the full dome, down any of a circle of directions.

No solution is preferred. The data do not contain $T$; the law does not produce it; and the state at time $0$ is identical in every case.

Compare the harmonic-oscillator non-example of section 1: there the ambiguity came from supplying half a state. Here the whole state is supplied.

what is not wrong

Energy: conserved, and equal to that of the resting mass, at every instant.
Continuity of the force: $F(s) = mk\operatorname{sgn}(s)\sqrt{|s|}$ is continuous, and $F(0)=0$.
The surface: smooth in the sense that matters — $h \in C^1$ — with a well-defined horizontal tangent plane at the apex. (It is not $C^2$ there: the curvature diverges.)
Newton's laws: every one is satisfied by every member.

Only one thing is missing, and section 2 says which.

The curvature claim, checked. $h'(r) = -(k/g)\sqrt r$ so $h''(r) = -(k/g)/(2\sqrt r) \to -\infty$ as $r \to 0^+$. So the meridian has unbounded curvature at the apex even though it has a perfectly good tangent there. Some readers locate the trouble here and say the dome is not a physical surface. It is a fair objection to the physics; it is not a defence of determinism, because the mathematical failure is a failure of a hypothesis about $f$, and one can ask the same question of any $\ddot x = k|x|^{1/2}\operatorname{sgn} x$ whatever its provenance.

Is this "the mass decides"? No. There is no mechanism, no hidden variable and no randomness in the model — the model simply does not determine the motion. The correct statement is that the initial value problem is not well posed, and any further story is added by the teller.

The standard replies, and what each concedes. (i) "Idealization": true, but so is every frictionless surface in mechanics, and the objection would delete most of Part A. (ii) "Measure zero": also true — the bad set is one point of phase space — and it concedes that determinism is a statement holding almost everywhere rather than everywhere. (iii) "Quantum mechanics fixes it": possibly, and irrelevant to what Newtonian mechanics asserts, which is the question.

01 · the dome

The missing hypothesis, named

the diagnosis

$f(s,v) = \big(v, k\operatorname{sgn}(s)\sqrt{|s|}\big)$ is continuous on $\mathbb{R}^2$ but not Lipschitz in $y$ on any neighbourhood of the origin, because $\dfrac{\mathrm{d}}{\mathrm{d}s}\sqrt{s} = \dfrac{1}{2\sqrt s}$ is unbounded as $s \to 0^+$. Picard–Lindelöf therefore does not apply at $(0,0)$ — and its conclusion is false there.

this is not a gap in the theorem

The dome shows the Lipschitz condition cannot simply be dropped. Continuity alone gives existence (next slide) and nothing more. The hypothesis is not a technical convenience; it is the entire difference between "a future" and "the future".

01 · the dome

A Hookean apex against the dome

A useful way to remember it: Lipschitz bounds the rate at which the force distinguishes nearby states. Where that rate is infinite, states that differ by nothing can be driven apart.

On the dome, an infinitesimal displacement from the apex is repaid with an acceleration infinitely larger in proportion — $k\sqrt{s} \gg s$ for small $s$ — and that is what lets motion start from rest.

Solid: a linear (Hookean) apex, $\ddot s = s$, released at rest from $s = \epsilon$. Each factor of $100$ closer to the top costs the same extra time, and from $\epsilon = 0$ the mass never leaves. Dashed: the dome, which leaves from exact rest.

The slogan, made honest. "$k\sqrt s \gg s$" is the statement that the force is superlinear in the displacement at the apex, i.e. $\lim_{s\to 0^+} k\sqrt s / s = \infty$. Compare a Hookean apex, $F = -m\kappa s$ or $F = +m\kappa s$: there the ratio is the constant $\kappa$, the field is Lipschitz, and the mass at the apex stays put (or, for the unstable sign, takes an infinite time to leave — the familiar inverted pendulum, which approaches the top only asymptotically). The dome replaces an exponential departure with a finite-time one.

The figure, computed. Dimensionless, $\kappa = k = 1$. The unstable linear apex $\ddot s = s$ from $s(0) = \epsilon$, $\dot s(0) = 0$ has $s = \epsilon\cosh t$, which reaches $s = 1$ at $t = \operatorname{arcosh}(1/\epsilon)$: $5.30$, $9.90$, $14.51$, $19.11$ for $\epsilon = 10^{-2}, 10^{-4}, 10^{-6}, 10^{-8}$ — steps of $\ln 100 = 4.61$. The time to leave grows without bound as $\epsilon \to 0$, and at $\epsilon = 0$ it is infinite: the solution is $s \equiv 0$, the only one, as Picard–Lindelöf says. The dome's member $s = (t - 2)^4/144$ (dashed) starts from $\epsilon = 0$ exactly and still leaves, reaching $s = 1$ at $t = 2 + \sqrt{12} = 5.46$.

01 · the dome

Continuity alone: Peano, and the sharp line

Peano existence theorem · cited

If $f$ is merely continuous on an open $\Omega \ni (y_0,t_0)$, the initial value problem has at least one solution on some interval $[t_0-a, t_0+a]$. Uniqueness is not claimed.

Proof by Euler polygons plus the Arzelà–Ascoli theorem, which this course does not develop: see Hartman, Ordinary Differential Equations, Thm II.2.1, or Teschl, Ordinary Differential Equations and Dynamical Systems, §2.7. Peano, 1890.

So failure mode 1 of section 1 never occurs in mechanics: continuous forces always have trajectories. Only failure mode 2 is in play.

Osgood's criterion · cited

Uniqueness already follows from $\|f(y,t)-f(z,t)\| \le \omega(\|y-z\|)$ with $\omega$ continuous, increasing, $\omega(0)=0$ and

$$\int_{0^+} \frac{\mathrm{d}u}{\omega(u)} = \infty .$$

Lipschitz is $\omega(u) = Lu$, whose integral diverges. The dome has $\omega(u) \sim k\sqrt u$, whose integral $\int_{0^+} u^{-1/2}\,\mathrm{d}u$ converges. The dome sits on the far side of a sharp line, not just outside a convenient hypothesis.

Why the Osgood integral is exactly the right quantity. Suppose two solutions leave the same point and $\varphi(t) = \|y(t)-z(t)\|$. Informally $\dot\varphi \le \omega(\varphi)$, so separating variables (unit 00's tool) gives $\int_{\varphi(t_0)}^{\varphi(t)} \mathrm{d}u/\omega(u) \le t - t_0$. If the integral from $0$ diverges, no finite time suffices to move $\varphi$ off $0$, and the solutions cannot separate. With $\omega(u) = ku^{1/2}$ the integral from $0$ is finite: $\int_0^\varepsilon \mathrm{d}u/(k\sqrt u) = 2\sqrt\varepsilon/k < \infty$, so separation in finite time is not forbidden — and the dome takes the opportunity. Osgood, 1898; the theorem itself is proved in Hartman, ch. III, where the uniqueness criteria are collected — this paragraph is the idea, not the proof.

Check the dome against the family. Two members with departure times $T$ and $T' > T$ separate, at time $t > T'$, by $\varphi(t) = \frac{k^2}{144}\big[(t-T)^4 - (t-T')^4\big]$, which is positive for every $t > T'$ however close $T'$ is to $T$. So $\int_{0^+}\mathrm{d}u/\omega(u)$ had better be finite, and with $\omega(u) = ku^{1/2}$ it is ($\int_0^\varepsilon \mathrm{d}u/(k\sqrt u) = 2\sqrt\varepsilon/k$). Consistent.

Peano is genuinely weaker. Its proof produces a solution as a limit of Euler polygons along a subsequence; different subsequences can converge to different solutions, and on the dome they do. There is no "canonical" choice hidden in the construction.

section 4

How far the guarantee reaches

Grönwall bounds how fast two futures separate; the maximal interval says when a future stops.

01 · reach

Grönwall's inequality

Let $u : [t_0,t_1] \to \mathbb{R}$ be continuous with $u \ge 0$, and let $A \ge 0$, $L \ge 0$ satisfy

$$u(t) \le A + L\int_{t_0}^{t} u(s)\,\mathrm{d}s \quad \text{for all } t \in [t_0,t_1]. \qquad\text{Then}\qquad u(t) \le A\,e^{L(t-t_0)}.$$

proof

Put $U(t) = A + L\int_{t_0}^t u$. Then $U$ is $C^1$ with $U' = Lu \le LU$, using $u \le U$ and $L \ge 0$. Hence

$$\big(U e^{-L(t-t_0)}\big)' = e^{-L(t-t_0)}\big(U' - LU\big) \le 0,$$

so $U(t)e^{-L(t-t_0)} \le U(t_0) = A$, giving $U(t) \le Ae^{L(t-t_0)}$, and $u \le U$. $\square$

checks

$L = 0$: reads $u \le A$, which is the hypothesis. ✓

$A = 0$: forces $u \equiv 0$ — an implicit inequality becomes an identity. That single line is a second proof of uniqueness.

Sharp: $u(t)=Ae^{L(t-t_0)}$ satisfies the hypothesis with equality, so the bound cannot be improved.

An implicit bound, where $u$ appears on both sides, is converted into an explicit one. That is the whole technique, and it recurs in units 02, 17 and 21.

Why $U' = Lu$ and not $Lu + {}$something: the fundamental theorem of calculus, since $u$ is continuous. And $u \le U$ is the hypothesis itself, restated. Those two facts are the entire proof; the exponential integrating factor is unit 00's separable-equation technique applied to the differential inequality.

The $A = 0$ case as a uniqueness proof. If $y,z$ solve the same equation with the same initial value and $f$ is $L$-Lipschitz, then subtracting the integral equations gives $\|y(t)-z(t)\| \le L\int_{t_0}^t\|y-z\|$, i.e. the hypothesis with $A = 0$; so $\|y - z\| \equiv 0$. This route needs no fixed-point theorem at all, but it gives uniqueness only — existence still needs Picard.

Sharpness, checked: for $u(t) = Ae^{L(t-t_0)}$, $A + L\int_{t_0}^t u = A + A(e^{L(t-t_0)}-1) = Ae^{L(t-t_0)} = u(t)$. Equality throughout.

Thomas Hakon Grönwall proved the differential form in 1919, in a paper on the dependence of solutions on a parameter; the integral form above is Richard Bellman's (1943), and both names are attached to it.

01 · reach

Continuous dependence on the initial state

continuous dependence

Let $f$ be $L$-Lipschitz in $y$ on a set containing both solutions, and let $y, z$ solve $\dot w = f(w,t)$ on $[t_0,t_1]$. Then

$$\|y(t) - z(t)\| \;\le\; \|y(t_0) - z(t_0)\|\;e^{L(t-t_0)} .$$

proof

Subtract the integral equations: $y(t)-z(t) = \big(y(t_0)-z(t_0)\big) + \int_{t_0}^{t}\big(f(y(s),s)-f(z(s),s)\big)\mathrm{d}s$. Take norms and use the Lipschitz condition: $\|y(t)-z(t)\| \le \|y(t_0)-z(t_0)\| + L\int_{t_0}^{t}\|y(s)-z(s)\|\,\mathrm{d}s$. Apply Grönwall with $u = \|y-z\|$ and $A = \|y(t_0)-z(t_0)\|$. $\square$

what it says, physically

A measurement error in the initial state grows at most exponentially, at a rate fixed by the force law and not by the trajectory. Halving the error in the data halves the bound at every later time.

So the solution map $y_0 \mapsto y(t)$ is continuous — indeed Lipschitz — which is the third condition in Hadamard's notion of a well-posed problem, after existence and uniqueness.

Check $L=0$: $\|y-z\|$ is constant, correct for $\dot y = f(t)$, where both solutions are the same function up to a constant shift.

Where the dome stands in this. It cannot even be stated there: the solution map is not a map, because one initial state has many futures. Continuous dependence presupposes uniqueness, and Grönwall's proof uses the Lipschitz constant that the dome does not have. This is the running example's appearance in section 4's first half — a theorem that has nothing to say about it.

The same estimate bounds the effect of changing the law. If $\dot y = f(y,t)$ and $\dot z = \tilde f(z,t)$ with $\|f - \tilde f\| \le \eta$ everywhere, the same argument gives $\|y(t)-z(t)\| \le \big(\|y(t_0)-z(t_0)\| + \eta (t-t_0)\big)e^{L(t-t_0)}$: perturb the data or perturb the force, the exponential is the same. Unit 21 lives on this estimate, and unit 10 uses it to say how much a planet's non-sphericity can move its satellite.

Hadamard's three conditions (existence, uniqueness, continuous dependence) were formulated for partial differential equations, where the third fails routinely — the backward heat equation is the standard example, and it will come back in unit 18.

01 · reach · example

How loose is the exponential?

worked example 4 — two pendulums, released a hair apart

Setup. Nondimensionalize with $\tau = t\sqrt{g/\ell}$, so the pendulum is $\ddot\theta = -\sin\theta$, $f(\theta,u) = (u,-\sin\theta)$, $L = 1$. Two masses are released from rest at $\theta_0 = 0.100$ and $\theta_0' = 0.101$ rad, a difference of $10^{-3}$.

The bound. Grönwall gives $\|\Delta y(\tau)\| \le 10^{-3}e^{\tau}$. At $\tau = 10$ that is $10^{-3} \times 22026 = 22.0$ rad — a useless statement, since the pendulum's angle never leaves $[-0.101, 0.101]$.

The truth. At these amplitudes $\sin\theta \approx \theta$ and the motion is $\theta(\tau) = \theta_0\cos\tau$, so $\Delta\theta(\tau) = 10^{-3}\cos\tau$ and $\|\Delta y(\tau)\| = 10^{-3}$, constant. The separation does not grow at all.

In seconds. For $\ell = 1$ m, $\sqrt{\ell/g} = 0.3193$ s, so $\tau = 10$ is $t = 3.193$ s, about $1.6$ periods of $2\pi\sqrt{\ell/g} = 2.006$ s.

the moral

Grönwall is a worst case over all equations with that Lipschitz constant, not a prediction for one of them. It is exactly sharp for $\dot y = Ly$ and wildly pessimistic for an oscillator.

Every number above was computed. $e^{10} = 22026.4658$, so the bound at $\tau = 10$ is $22.0265$ rad. $\sqrt{1/9.80665} = 0.319330$ s, so $\tau = 10$ is $t = 3.19330$ s and the small-amplitude period is $2\pi\sqrt{\ell/g} = 2.00641$ s, giving $3.19330/2.00641 = 1.592$ periods.

Honesty about the linearization. $\theta(\tau)=\theta_0\cos\tau$ solves $\ddot\theta = -\theta$, not $\ddot\theta = -\sin\theta$. What matters here is not how far each pendulum is from its linearization — that error is common to both and cancels in the difference — but that the two have slightly different periods: by unit 00's expansion the relative difference is $(\theta_0'^2-\theta_0^2)/16 = 1.26\times10^{-5}$. Over $\tau = 10$ that is a phase drift of about $1.3\times10^{-4}$ rad, which rotates the difference vector $\Delta y$ in the plane rather than lengthening it. The exact nonlinear motions have $\|\Delta y(10)\| = 9.9947\times10^{-4}$ against the linearized $10^{-3}$: constant to four figures, and growing at most linearly thereafter. The bound is exponential, the truth is flat: the moral stands.

01 · reach · example

The bound against the truth, plotted

Worked example 4 on a logarithmic axis: the separation of the two pendulums against the Grönwall bound, over $0 \le \tau \le 10$.

and yet

Some systems saturate it. When the true separation really does grow like $e^{\lambda \tau}$ with $\lambda > 0$, the exponent is a Lyapunov exponent and the system is chaotic — deterministic, and unpredictable. Unit 17 measures $\lambda$; this unit only proves it cannot exceed $L$.

Log scale. Dashed: the Grönwall bound $10^{-3}e^{\tau}$, which $\dot y = y$ attains exactly. Solid: the two pendulums, computed — flat at $10^{-3}$, while the bound reaches $22$ rad.

01 · reach

The maximal interval of existence

maximal interval of existence

Under the hypotheses of Picard–Lindelöf on an open $\Omega$, every initial value problem has a unique solution on an open interval $(T_-, T_+) \ni t_0$ that extends every other solution.

escape lemma

If $T_+ < \infty$ then the solution leaves every compact subset of $\Omega$: for each compact $K \subseteq \Omega$ there is $t < T_+$ with $(y(t),t) \notin K$.

So a solution can only end by running to the boundary of $\Omega$ or to infinity. It cannot simply stop.

why the maximal interval is open

If the solution existed at $T_+$ itself, with $(y(T_+),T_+) \in \Omega$, the theorem applied at that point would extend it past $T_+$, contradicting maximality.

Existence: take the union of all solutions through $(y_0,t_0)$. Global uniqueness makes the union a well-defined function, and it solves the equation because solving is a local property.

full proof of the escape lemma in the notes (N)

Full proof of the escape lemma

Suppose $T_+ < \infty$ and $(y(t),t) \in K$ for all $t \in [t_0,T_+)$, with $K \subseteq \Omega$ compact. Then $\|f\| \le M_K := \max_K\|f\|$ along the solution, so for $t < t' < T_+$, $\|y(t')-y(t)\| = \big\|\int_t^{t'} f\big\| \le M_K|t'-t|$: the solution is Lipschitz in $t$, hence uniformly continuous, hence has a limit $y_\ast = \lim_{t\to T_+^-} y(t)$ (Cauchy criterion). The point $(y_\ast,T_+)$ lies in $K$ because $K$ is closed, so it lies in $\Omega$. Extend $y$ to $[t_0,T_+]$ by $y(T_+) = y_\ast$; the extension is continuous and satisfies the integral equation up to $T_+$ (let $t \to T_+$ in it), hence solves the ODE there. Now apply Picard–Lindelöf at $(y_\ast,T_+)$ to get a solution on $[T_+-a, T_++a]$; by global uniqueness it agrees with $y$ where both are defined, so together they form a solution past $T_+$. That contradicts maximality. $\square$

The two ways out, in the two standard settings. If $\Omega = \mathbb{R}^N\times\mathbb{R}$, compact means closed and bounded, so escaping every compact set with $T_+$ finite means $\|y(t)\| \to \infty$: finite-time blow-up. If $\Omega$ is a proper open subset — the dome, with $|s| < g^2/k^2$ — the solution may instead reach the boundary in finite time while staying bounded.

Note what is not claimed: that $y(t)$ has a limit as $t \to T_+$. For blow-up it need not; $\|y(t)\|$ can oscillate while growing. The clean statement is the one about compact sets.

01 · reach · example

Two solutions that end

worked example 5a — $\dot x = x^2$, a model equation

Step 1. Separate (unit 00): $\int x^{-2}\mathrm{d}x = \int \mathrm{d}t$ gives $-1/x = t + C$.

Step 2. With $x(0) = x_0 > 0$, $C = -1/x_0$ and $\;x(t) = \dfrac{x_0}{1 - x_0 t}$.

Step 3. $x(t) \to \infty$ as $t \to 1/x_0$ from below, so $T_+ = 1/x_0$. For $x_0 = 2$: the solution ceases to exist at $t = 0.5$.

Step 4. Yet $f(x) = x^2$ is as smooth as anything, and locally Lipschitz on all of $\mathbb{R}$. No hypothesis failed; the theorem never promised more than local existence.

worked example 5b — an inverse-cube attractor

Step 1. $m\ddot x = -k/x^3$ on $x > 0$, released from rest at $x = a$ ($k$ and $a$ are new constants here, not the dome's $k$ or the theorem's $a$). Unit 00's third case: $V(x) = -k/(2x^2)$ and $E = -k/(2a^2)$.

Step 2. $\tfrac12 m\dot x^2 = E - V$ gives $\dot x = -\sqrt{\tfrac{k}{m}}\,\dfrac{\sqrt{a^2-x^2}}{a\,x}$ (taking the inward root).

Step 3. Quadrature: $\;t_{\text{fall}} = \sqrt{\tfrac{m}{k}}\,a\displaystyle\int_0^a \frac{x\,\mathrm{d}x}{\sqrt{a^2-x^2}} = \sqrt{\tfrac{m}{k}}\,a\Big[-\sqrt{a^2-x^2}\Big]_0^a = a^2\sqrt{\tfrac{m}{k}}$.

Step 4. Finite. With $a = 2$ m, $m = 0.5$ kg, $k = 1\ \mathrm{kg\,m^4\,s^{-2}}$: $t_{\text{fall}} = 4\sqrt{0.5} = 2.828$ s, and the speed diverges on arrival.

01 · reach · example

Two endings, plotted

The escape lemma says a solution with $T_+ < \infty$ leaves every compact subset of $\Omega$. The two examples are its two ways out.

5a has $\Omega = \mathbb{R}\times\mathbb{R}$: the only way out is up, and $x(t) = x_0/(1 - x_0 t)$ goes to infinity at $t = 1/x_0$.

5b has $\Omega = \{x > 0\}$: the solution $x(t) = \sqrt{a^2 - kt^2/(ma^2)}$ stays bounded and reaches the edge $x = 0$, arriving with infinite speed.

checks

$[k] = \mathrm{M\,L^4\,T^{-2}}$, so $a^2\sqrt{m/k}$ is a time ✓; $a \to 0$ gives $t_{\text{fall}} \to 0$ ✓. Neither contradicts Picard–Lindelöf: 5a leaves every compact set of $\mathbb{R}$, 5b every compact subset of $\Omega = \{x > 0\}$.

Left, 5a: $x_0/(1 - x_0t)$ for $x_0 = 2$ and $1$, escaping to infinity at $t = 1/x_0$. Right, 5b: the fall from $a = 2$ m, reaching the edge $x = 0$ of $\Omega$ at $2.83$ s with vertical slope — infinite speed.

5b in closed form. From Step 3 with the upper limit left free, the time to fall from $a$ to $x$ is $t(x) = \sqrt{m/k}\,a\int_x^a \frac{x'\,\mathrm{d}x'}{\sqrt{a^2 - x'^2}} = \sqrt{m/k}\,a\sqrt{a^2 - x^2}$, so $x(t) = \sqrt{a^2 - kt^2/(ma^2)}$: a quarter of an ellipse in the $(t, x)$ plane, with vertical tangent where it meets $x = 0$ at $t = a^2\sqrt{m/k} = 2.828$ s. The right panel plots it for the slide's numbers, $x = \sqrt{4 - t^2/2}$.

What "ceases to exist" means physically for 5b. Nothing in the model describes the mass after $t_{\text{fall}}$: the force law $-k/x^3$ is not defined at $x = 0$, and the kinetic energy is unbounded. The model has told you it is out of its depth. The inverse-square law, by contrast, gives a finite arrival speed only because $V = -k/x$ is less singular — and even there the two-body problem's collision is a genuine end of the solution, which unit 08 will have to handle.

01 · reach

When does a solution live forever?

global existence

Let $f$ be continuous and locally Lipschitz on $\mathbb{R}^N\times\mathbb{R}$, and suppose that on each strip $\mathbb{R}^N \times [t_0,t_1]$ there is $C$ with $\;\|f(y,t)\| \le C\big(1 + \|y\|\big)$. Then every solution exists on all of $\mathbb{R}$.

proof

On $[t_0,t_1]$, $\ \|y(t)\| \le \|y_0\| + \int_{t_0}^{t} C(1+\|y\|) \le \big(\|y_0\| + C(t_1-t_0)\big) + C\int_{t_0}^{t}\|y\|$. Grönwall with $A = \|y_0\|+C(t_1-t_0)$, $L = C$ gives $\|y(t)\| \le Ae^{C(t-t_0)}$, a bound depending only on $t_1$. So the solution stays in a fixed closed ball, a compact set, on $[t_0,t_1)$; by the escape lemma $T_+ > t_1$. As $t_1$ was arbitrary, $T_+ = \infty$; likewise $T_- = -\infty$. $\square$

the pendulum is deterministic forever

$f(\theta,u) = (u, -\sin\theta)$ has $\|f\| \le \sqrt{u^2 + 1} \le 1 + \|y\|$, so every pendulum motion exists for all time and is unique. Determinism, for this system, is a proved theorem — not an assumption.

Why 5a is not covered: $x^2$ grows faster than $C(1+|x|)$ for any $C$. The hypothesis is precisely what rules out blow-up, and it is close to sharp: $\dot x = x^{1+\varepsilon}$ blows up for every $\varepsilon > 0$.

Linear growth, not boundedness, is the right hypothesis. A bounded $f$ obviously gives global existence; but the most important systems in this course — linear ones, $\dot y = Ay$ (unit 07) — have unbounded $f$ and satisfy $\|Ay\| \le \|A\|\,\|y\|$. Linear growth covers both, which is why it is stated this way.

Check the pendulum bound. $\|f(\theta,u)\|^2 = u^2 + \sin^2\theta \le u^2 + 1$, and $\sqrt{u^2+1} \le 1 + |u| \le 1 + \|y\|$. So $C = 1$ works on every strip.

A second route for mechanics, worth knowing. If the energy $E = \tfrac12 m\|\dot x\|^2 + V(x)$ is conserved and $V$ is bounded below, then $\|\dot x\|$ is bounded by $\sqrt{2(E - \inf V)/m}$ and $\|x\|$ grows at most linearly, so the solution stays in a compact set on any finite interval and lives forever. That covers the pendulum, every bounded potential, and the harmonic oscillator, and it fails exactly for potentials unbounded below — like 5b's $-k/2x^2$, which is why 5b can end.

01 · the dome, closed

How long does a dome solution live?

The dome's domain is $\Omega = \{(s,v) : |s| < g^2/k^2\}\times\mathbb{R}$ — the surface simply ends where it turns vertical. Take the member $r_T$ of the family.

its maximal interval

$T_- = -\infty$: the solution is $\equiv 0$ into the past, forever. ($\equiv 0$ is a past, not the past: the same failure runs backwards.)

$T_+ = T + \big(144\,g^2/k^4\big)^{1/4}$, when $r$ reaches $g^2/k^2$. With $k = 1\,\mathrm{m^{1/2}s^{-2}}$: $\;T_+ - T = (144 \times 96.17)^{1/4} = \mathbf{10.85}$ s.

At that instant the solution leaves every compact subset of $\Omega$ without going anywhere near infinity — it simply arrives at the boundary. Escape lemma, second mode.

The departing member, $k = 1$. It reaches $r = 57.70$ m at $9.55$ s, where the mass leaves the surface, and the rim, $r = 96.17$ m, at $10.85$ s, where $\Omega$ ends.

The formula for $T_+$. Solve $\frac{k^2}{144}(T_+-T)^4 = \frac{g^2}{k^2}$: $(T_+-T)^4 = \frac{144 g^2}{k^4}$, so $T_+ - T = \big(144 g^2/k^4\big)^{1/4} = \frac{12^{1/2}\sqrt{g}}{k}$. Check the dimensions: $[\sqrt g / k] = (\mathrm{L\,T^{-2}})^{1/2}/(\mathrm{L^{1/2}T^{-2}}) = \mathrm{T}$. ✓ Numerically $\sqrt{12}\sqrt{9.80665}/1 = 3.4641\times 3.1316 = 10.848$ s, agreeing with $(144\times96.17)^{1/4}$.

Reading the figure. $r = (t - T)^4/144$ in metres, against the time since departure. Solid up to the moment the mass leaves the surface, dashed where only the mathematics continues; the dashed curve hits the rim, the boundary of $\Omega$, at $10.85$ s, and the solution of the ODE ends there.

A coincidence worth noticing: the dome is exactly as deep as its rim is far from the axis, both equal to $2g^2/(3k^2) = 64.11$ m (so it is $128.2$ m across). The depth is $\tfrac{2k}{3g}(g^2/k^2)^{3/2}$ and the horizontal radius is $\int_0^{g^2/k^2}\sqrt{1-(k/g)^2 r}\,\mathrm{d}r$; both come to $2g^2/(3k^2)$.

01 · the dome, closed

The physical solution ends first

and the mass leaves earlier still

The normal force vanishes at $r = 57.70$ m, reached at $t - T = 9.55$ s at a speed of $24.17\ \mathrm{m\,s^{-1}}$. After that the mass is a projectile and the dome's equation no longer models it. The mathematical solution ends at $10.85$ s; the physical one at $9.55$ s.

Three different reasons a solution can stop — blow-up, boundary of $\Omega$, and the model ceasing to apply — and only the first two are theorems. Telling them apart is part of the job.

Normal force along the departing solution, against arc length. It starts at $mg$ (the mass on a horizontal tangent plane), falls through zero at $57.70$ m and would go to $-\infty$ at the rim.

The normal force. Along the family, energy gives $\dot r^2 = -2gh(r) = \tfrac{4k}{3}r^{3/2}$, independent of $T$. With $u = (k/g)\sqrt r$, the component of Newton's law along the meridian's normal gives $N = mg\big(\sqrt{1-u^2} - \tfrac23 u^2/\sqrt{1-u^2}\big)$ — derived in full in the notes to "Norton's dome: the surface, and its equation" — zero at $u^2 = 3/5$, i.e. $r = \tfrac35 g^2/k^2 = 57.70$ m, where the surface tilts at $\arcsin\sqrt{3/5} = 50.77^\circ$. The time follows from $(144\times 57.70)^{1/4} = 9.547$ s and the speed from $\sqrt{\tfrac43 k\,r^{3/2}} = 24.17\ \mathrm{m\,s^{-1}}$.

Reading the figure. The dashed curve is $\sqrt{1 - u^2}$, the cosine of the tilt: the normal force the surface would need if the mass were at rest there. The solid curve subtracts the curvature term $\tfrac23 u^2/\sqrt{1-u^2}$, the extra the surface would have to supply to bend a moving mass round a surface curving away from it. It crosses zero at $u^2 = 3/5$, i.e. $r = 57.70$ m, and beyond that a surface would have to pull, which a surface cannot do.

01 · summary

What determinism buys, and what it costs

you may now say

the solution, not a solution, wherever $f$ is locally Lipschitz in $y$;
a verified guess is the answer — the working method of units 06, 07, 08;
an error in the data costs at most $e^{L\Delta t}$, so approximations mean something;
a solution that ends, ends by escaping: to infinity, or to the boundary.

and you may not say

that determinism is automatic — one continuous force law refutes it;
that a smooth law gives solutions for all time — $\dot x = x^2$ refutes it;
that determinism gives predictability — the bound is exponential, and unit 17 shows it is sometimes attained;
that Lipschitz is a technicality — Osgood puts the dome on the far side of a sharp line.

Next: unit 02 asks what you can say about the solution when there is no formula for it, and the answer is drawn in the phase space this unit built.

01 · checkpoint

Done when

checkpoint

You state and prove the Picard–Lindelöf theorem — the integral equation, the tube, the weighted norm, the contraction, the fixed point — and you exhibit Norton's dome's one-parameter solution family together with the exact hypothesis that fails, without appealing to physical intuition at any step.

How to fail it, unambiguously

Any of: the statement omits the cylinder or the interval $\min(c, b/M)$; the proof does not say where completeness is used; the family is displayed without verifying it is $C^2$ at $t = T$; or the diagnosis says "the force is not smooth" rather than naming the Lipschitz condition and the point where it fails.

Then

./mr problems 01 — P10 is the proof, P11 the dome, P9 the blow-up.
Hints, then the worked solution, fold under each problem — only after an attempt.
./mr done 01 when the checkpoint is met.

Reading, all optional: Teschl §§2.2–2.6; Hartman ch. II–III; Arnold, Ordinary Differential Equations, ch. 2; Norton (2008) for the dome.