← Vol. IV Home Mini-Curso · Session S1 / 3 S2: Theorem & Basin →
MINI-CURSO · SESSION S1 · 60 MIN
S1

Contact Geometry & the dm³ System

The arena: a contact manifold in dimension 3, the Reeb field, and the dissipative ODE whose attractor is a helix. The geometry forces it.
Objective: introduce contact geometry by worked example | Pre-req: undergraduate ODE | Mode: lecture + live simulator
S1 · CONTACT GEOMETRY  →  S2 · THEOREM 2.1 & BASIN  →  S3 · LEAN 4 SKELETON
Session Agenda · 60 Minutes

What We Will Cover

  1. 0 – 12 min — Contact manifolds. Why $\alpha = dz - r^2\,d\theta$ on $\mathbb{R}^3$ is the prototype.
  2. 12 – 25 min — The Reeb vector field. Why non-integrability of $\xi = \ker\alpha$ turns periodic orbits into helices.
  3. 25 – 40 min — The dm³ ODE: Hopf normal form with a contact-aware coupling $\varepsilon(r-1)e^{-z}$.
  4. 40 – 55 min — Live simulator. Trajectories from many $r(0)$; what spirals in, what falls out.
  5. 55 – 60 min — Hand-off to S2: the symmetric $\varepsilon_0 = 1/3$ Gronwall ball we will later correct.
§1 · Foundation

What Is a Contact 3-Manifold?

Classical dynamics likes symplectic geometry — even-dimensional phase spaces, flows that preserve area. But living systems, weather, the circadian clock: they all happen in odd dimensions and they all rotate. Contact geometry is the odd-dimensional cousin of symplectic geometry. It is the natural setting for dissipative rotating flows.

Definition 1.1 · Contact 3-Manifold
A contact 3-manifold is a pair $(M, \xi)$ where $M$ is a smooth 3-manifold and $\xi$ is a totally non-integrable plane field. "Non-integrable" is captured by a 1-form $\alpha$ with $\xi = \ker\alpha$ and $$\alpha \wedge d\alpha \;\ne\; 0 \quad \text{everywhere on } M.$$ The form $\alpha$ is called a contact form. Non-integrability means: at no point does a small 2-dimensional integral surface exist tangent to $\xi$. Geometrically: the planes twist so violently that they refuse to assemble into surfaces.

The prototype: $\mathbb{R}^3$ with $\alpha = dz - r^2\,d\theta$

Use cylindrical coordinates $(r, \theta, z)$ on $\mathbb{R}^3 \setminus \{r = 0\}$. Define

$\alpha \;=\; dz \;-\; r^2\,d\theta.$

Compute $d\alpha = -2r\,dr\wedge d\theta$, so

$\alpha \wedge d\alpha \;=\; (dz - r^2\,d\theta)\wedge(-2r\,dr\wedge d\theta) \;=\; -2r\,dz\wedge dr\wedge d\theta \;\ne\; 0.$

This proves $(\mathbb{R}^3, \ker\alpha)$ is a contact 3-manifold. The planes $\xi$ rotate around the $z$-axis by an angle that grows with $r$ — they "overtwist" away from vertical.

§2 · The Reeb Flow Forces Helices

The Reeb Vector Field

Definition 2.1 · Reeb Field
Given a contact form $\alpha$ on $M$, the Reeb vector field $R$ is the unique vector field satisfying $$\alpha(R) = 1, \qquad d\alpha(R, \cdot) = 0.$$ It is the natural "time direction" on $(M, \xi)$ — transverse to every contact plane, and its flow preserves $\alpha$.

For the prototype $\alpha = dz - r^2\,d\theta$, one checks $R = \partial_z$. The Reeb flow is therefore vertical translation: $\Phi_t(r,\theta,z) = (r,\theta,z+t)$. That might look disappointing — but we want the flow of the full contact system, not just the Reeb flow of a specific $\alpha$. The model orbit we care about is the one that lives on the cylinder $r = 1$ and spirals uniformly: $\dot\theta = 1, \dot z = 1$. That is the helix.

Claim · Helicity Is Forced
Periodic planar orbits cannot lift to $\mathbb{R}^3$ contact dynamics without becoming helical.
Given the non-integrability $\alpha\wedge d\alpha \ne 0$, any closed trajectory in a plane transverse to $\xi$ must have a non-zero $dz$ component per revolution. In the prototype, this is exactly $\oint dz = \oint r^2\,d\theta = 2\pi r^2 \ne 0$. One revolution in $\theta$ forces exactly $2\pi r^2$ units of vertical climb. This is the geometric reason the attractor below is a helix rather than a circle.
§3 · The Main Character

The dm³ System

We study the autonomous system on $M = \mathbb{R}^3$ (in cylindrical coordinates) with fixed coupling $\varepsilon = 2$:

$\dot r \;=\; r(1 - r^2) \;+\; \varepsilon(r - 1)\,e^{-z}, \qquad \dot\theta \;=\; 1, \qquad \dot z \;=\; r^2 \;-\; \varepsilon(r - 1)^2\,e^{-z}.$
Equation (1) · the dm³ ODE · preprint §2

Anatomy of the radial equation

Anatomy of the vertical equation

On the attractor $r=1$ the coupling term vanishes and $\dot z = r^2 = 1$: vertical velocity is exactly unity. Combined with $\dot\theta = 1$, trajectories on $r = 1$ are parametrised by $(\cos t, \sin t, t)$ — the unit helix. This is our $\Gamma$: the Reeb orbit of $\alpha$ restricted to $r=1$.

$\Gamma = \{(r,\theta,z) : r = 1,\; \dot\theta = \dot z = 1\}$ is not a theoretical convenience. It is forced by three facts: (i) the Hopf term pins $r=1$; (ii) the contact coupling vanishes there; (iii) the contact form insists $\dot z$ equals $r^2$ on the limit set. All three independently point to the same object. The attractor could not be anything else.

Why $\varepsilon$ matters

Set $\varepsilon = 0$. The system decouples: $(r,\theta)$ follows Hopf in the plane, $z = t$ rides along. $r=1$ is a normally hyperbolic invariant manifold and standard theory gives global convergence from $r(0) > 0$. Boring.

Now set $\varepsilon = 2$. The coupling term is asymmetric in $r$ (linear in $r-1$) but symmetric in sign of deviation — and it grows exponentially when $z$ is negative. That asymmetry/exponential combination is the source of the pedagogical surprise we will expose in S2: the basin of attraction of $\Gamma$ is not a symmetric neighbourhood of $r=1$.

Foreshadowing S2: A Gronwall estimate will give us a symmetric ball $|r-1| < 1/3$ of guaranteed convergence. The numerics will show the real inner boundary is at $r \approx 0.80$, not $r = 2/3$. The symmetric estimate is not just coarse — it misclassifies orbits. This is the core pedagogical contribution of the course.
§4 · Play With It

Live Simulator — See the Helix Form

▶ dm³ Contact Flow — 3D View

Drag to rotate · scroll to zoom · slide $r_0$ and $\varepsilon$. Blue trajectories spiral in from $r(0) > 1$. Red trajectories from too-small $r(0)$ escape downward when $e^{-z}$ explodes.

Guided experiments for the 15-minute segment

  1. Confirm the helix. Set $r_0 = 1.0$, $\varepsilon = 0$. Watch the trajectory hug $r=1$ while climbing. Increase $\varepsilon$ to $2$ — the trajectory still lands on the helix; the coupling does not disturb $\Gamma$ itself.
  2. Outer basin. Try $r_0 \in \{1.1, 1.5, 2.5\}$ with $\varepsilon = 2$. All converge. Note how quickly they approach $r=1$.
  3. Inner trap. Try $r_0 = 0.9, 0.8, 0.7$. Something interesting happens between 0.9 and 0.7. You will feel the asymmetric boundary before you prove it exists.
  4. Read out $\mu$. Notice the rate $\hat\mu$ converges to a number near $-2$ as $r$ approaches $1$. This is the key observation for S2.
§5 · What's Next

Hand-off to Session S2

You have now seen:

In S2 we prove the exponential convergence rigorously: linearisation gives eigenvalue $-2$, Gronwall gives a symmetric ball $|r-1| < 1/3$ of guaranteed contraction, and the high-precision DOP853 integrator gives us Table 1 — which corrects the inner boundary to $r_* \approx 0.80$. The correction is the pedagogical heart of the course.

Board exercise · Take home for S2
Verify $\{r > 1\}$ is positively invariant.
Show that along solutions of (1), if $r(0) > 1$ then $r(t) > 1$ for all $t \geq 0$. Hint: compute $\dot r$ at $r = 1$. The sign of the contact coupling at $r = 1$ is zero; what does $r(1-r^2)$ contribute at $r = 1 + \delta$ for small $\delta > 0$? This invariance is what lets the Gronwall argument run without a sign ambiguity.
The outer basin is easy. The inner boundary is hard. That is the whole course in one sentence. Carry it into S2.