Is a reasonable assumption to consider that the contact point of the Euler's Disk (with stationary center of mass) trace this finite bounded spiral?

This question is highly related to working with the assumption of systems achieving a finite ending time, which is better explained in this other question.


After seeing this videos about the mathematics of the Euler's Disk toy: [video 1], [video 2], and [video 3], I am trying to understand the dynamic of the Euler's Disk when it behaves like having a finite-time blow up for its wobbling rate.

I started from the paper "Euler's Disk and its finite-time singularity" by H. K. Moffatt (2000) [paper 1], and there is presented that $\omega = -\Omega\sin(\alpha)$ and $\Omega^2\sin(\alpha) = \frac{4g}{a}$, and then through an analysis through approximations leads to a result for $\approx \alpha(t) \rightarrow \,\approx \Omega(t)$ showing the finite time blow up approach (which can be seen directly indeed since a the time the motion stop $\alpha \to 0$ so $\Omega = \sqrt{\frac{4g}{a\sin(\alpha)}}$:

Image from paper 1

But since there is no deep explanation of the maths, I move to this other paper "The Rolling Motion of a Disk on a Horizontal Plane" by A. J. McDonald & K. T. McDonald (2001) [paper 2] where the differential equations of motions are displayed.

I am still not sure if I mess it up pairing constants and variables among the videos and the papers, so I will try to use it the less possible to focus on the main topic of the question.

From paper 2, on equations (23) to (25) there are the equations of motion of the Euler's Disk, but unfortunately are not enough information to find $\alpha(t)$ neither a differential equation only dependent on $\alpha$ and its derivatives (at least I wasn't able to do it). On paper 2, on the sections of "Steady motion", it is again found the result $\Omega_0^2\sin(\alpha_0) = \frac{g}{ka}$ (with $k=1/4$ for a disk). Using this result again into Eqs. (23), (24), or (25) only works to find nonsense results for $\alpha(t)$ (if any).

Here I realize that I am thinking the problem mistakenly: been right the assumption that the center of mass its stationary (is not true in the real world in general, but it could be done for small angles - it match the condition $b=0$ for paper 2), the results of paper 1 and the Steady Motion of paper 2 are considering that the contact point of the Euler's disk it making circles on the flat surface where it spins, so in every revolution the center of mass is not changing its position (like in the amazing set-up experiment it is shown on video 2).

From now on, taking the Assumption 1: that the center of mass of the disk only moves on the vertical line, so every time the disk is making a revolution the center of mass has to get lower by a little bit, so, the disk has extended a little bit flatter, so on each revolution the projection of the radius on the flat surface (referred from the vertical line where the center of mass is moving) has to get a bit longer... so under this assumption, the point of contact between the disk and the flat surface must be doing some kind of Spiral.

Since the extended radius which is doing the spiral can achieve at maximum the value of the radius of the disk, it suggest that at the end of the movement the distance between loops must be reducing, so at first I think of things like the Fermat's Spiral, but then I realize it doesn't achieve a maximum value for the radius so instead I must consider a Bounded Spiral, as the example show on Wikipedia, for $(r,\,\varphi)$ the 2D polar coordinates: $r(\varphi)=a\tan^{-1}(q\varphi)$. But then, I realized that this last example will only achieve their final value at time $t \to \infty$, so If I believe that the disk indeed has stop moving, I need to move with other kind of functions.

Main "issues"

As in the situation of a variable going to zero at a finite time $T$: $y(T)=0$, will imply that its reciprocal $z(t)=\frac{1}{y(t)}$ will diverge because of having a singularity at $t=T$: $\lim_{t \to T} |z(t)| \to \infty$, this is why I believe that the singularity shown on the wobbling rate of the Euler's Disk is just an indication that the real world object have stop moving (I am counting the number of full turns per rotation, and rotations have become zero), so, I am using from now on Assumption 2: the Euler Disk stops moving at a finite time $0<T<\infty$... and in this assumption is where the "issues" begins.

Considering now $r(t)=a\cos(\alpha)$ the radius of the spiral is making, somehow I need to find a function $r(t)$ which: (i) achieve the value zero at finite time, (ii) could be described with a differential equation, so I start looking for related things in the web: "finite-time", "finite-time-convergence", "finite-duration", "time-limited", "compact-supported time", among others, and I found very little information (and very abstract), and also since there is no general name to this kind of functions I will defined here as:

Definition 1 - Solutions of finite-duration: they becomes exactly zero at a finite time $T<\infty$ by its own dynamics and stays there forever after. So, they are different of just a piecewise section of any arbitrarily function multiplied by rectangular function: it must solve the differential equation in the whole domain. (Here I just pick the shorter name its look more natural to me).

Since the Euler's Disk is rolling unperturbed with any kind of feedback, I focus on solutions to scalar Autonomous (so time Invariant) Ordinary Differential Equations (ODEs) for simplicity, instead of Partial Differential Equations (PDEs). I found these papers from Vardia T. Haimo (1985): "Finite Time Differential Equations" [paper 3] and "Finite Time Controllers" [paper 4], where there is proved that indeed exists this kind of solutions of finite-durations to ODEs (but unfortunately, no exact solution example is given). Caution here since in paper 3 there is an important sign mistake in an inequality so I recommend you to start from paper 4 (are very similar, and the last one is more easy to read), but there is an important issue mentioned on paper 3 which I would like to share here explicitly: without loss of generality, the papers considers that $T=0$ and that the scalar autonomous ordinary differential equations are at least class $C^1(\mathbb{R}\setminus\{0\})$:

"One notices immediately that finite time differential equations cannot be Lipschitz at the origin. As all solutions reach zero in finite time, there is non-uniqueness of solutions through zero in backwards time. This, of course, violates the uniqueness condition for solutions of Lipschitz differential equations."

This could be translated into these three facts:

  • FACT 1: no scalar linear ODE could support finite duration solutions!
  • FACT 2: If I want to accurately model a finite duration phenomena through an scalar autonomous ODE, it must be Non-Lipschitz at least at one point
  • FACT 3: Non-uniqueness of solutions arise when working with finite duration solutions.

Additionally to this issues where now I know that solutions are nonlinear and don't stand uniqueness (big issue since almost everything I have read about differential equations supports its analysis on uniqueness), searching for functions that becomes zero after a finite time $t=T$, I found that if an analytic function is zero in a non-zero measure compact set of points, then by the Identity Theorem the function must be the zero function in the whole domain, so:

  • FACT 4: No real-valued analytic function could represent a finite duration solution on the whole $\mathbb{R}$ domain.

These 4 facts rules out everything I learned about differential equation in engineering, since now I understand that no standard real-valued Power Series could support a finite duration solution on the whole $\mathbb{R}$ domain.

Fortunately, the same Wikipedia page for analytic function show one kind of functions that could be an alternative solution: bump functions $\in C_c^\infty$, which are an example of Non-Analytic Smooth Functions which have sections where they behave as Flat functions, so they smoothly becomes zero as intended: as shown on the Wiki pages, it has also another issue, since the functions are smoothly matching the zero constant at its domain's edges $\partial t$ all its derivatives must become zero there $\lim\limits_{t \to \partial t^{\pm}} \frac{d^n}{dt^n}f(t) = 0$, so in these neighborhoods the function cannot be represented through a Taylor's Series, since at these "ending points" all their coefficients will be zero, matching only the representation of the zero function (this is why are non-analytic).

I believe this last issue means also that the small-angle approximations cannot accurately model the behavior of a finite-duration solution at its ending time, since is equivalent to using the first coefficients of a Taylor's Series to approximate the function in a point it is not analytic (let's call this FACT 5, but I am not 100% sure is a fact).

Now, testing a classical smooth bump function as the main component of the increasing radius of the bounded spiral as $r(t) = 1-\exp(t^2/(t^2-1))\theta(1-t),\,t\geq 0$ with $\theta(t)$ the standard unitary step function (Heaviside step function), as is plot here, being a valid alternative in principle, these kind of functions shows to have 2 problems:

  • Since all the smooth bump functions I found here shows to behave as having a flat-top (or near-flat) at the beginning $t=0$, and a flat-end when reaching the edges of its compact support, the bounded spirals behaved as having a wider "pitch" at the middle than at the beginning, which doesn't look right to me (at least for the Euler's Disk problem under assumptions 1 and 2).
  • I wasn't able to find any smooth bump function which is described by a known scalar autonomous ODE, which I have asked here Examples of Finite-Duration solutions to Autonomous Ordinary Differential Equations ODEs?.

Luckily, in this answer to the just mentioned question some user (@Wrzlprmft) mentioned the existence of the Norton's Dome example, given by the differential equation $\ddot{r}=\sqrt{r}$ (dimensionless version). This is an example where non-uniqueness of solutions problems will rise: the initial conditions $r(0)=\dot{r}(0)=0$ are problematic since the solution $r(t)=\frac{1}{144}(t-T)^4$ will rise from zero at some arbitrary constant $T$ which is not dependent on the initial conditions, since the solution is indeed behaving as $r(t)=\frac{1}{144}(T-t)^4\theta(t-T)$ so its been already zero for any value of $0\leq t<T$ (see details here and here).

But fortunately, this equation is also an example of a differential equation that support finite duration solutions, since $r(t)=\frac{1}{144}(T-t)^4\theta(T-t)$ solves the differential equation in the whole domain $\mathbb{R}$. Even so, if I keep the initial value problem starting at $t=0$, the initial condition positive $r(0)>0$, and I focus only in real-valued solutions, the initial conditions $r(0)$ and $r'(0)$ will uniquely determined the ending time $T>0$ so I can pick only one possible solution (as when uniqueness is hold), so someway the finite duration solutions behave as the "well-behaved twin" of the Norton's Dome solutions.

After a lot of questions on MSE under the tag [finite-duration], I believe now if an scalar autonomous ODEs have a time $T>0$ where $r(T)=\dot{r}(T)=0$ and it also stands the trivial zero solution, then the differential equation could support the finite duration solutions $y(t) \equiv x(t)\theta(T-t)$ for $x(t)$ a non-trivial solution valid at least for $0\leq t<T$: see details here, here, and here.

Even if this finally ends to be untrue in general, fortunately I found this family of Non-Lipschitz ODEs where it can be showed it works properly for real-valued $n>1$, so I want to use them to model the radius of the bounded spiral: $$\dot{x} = -\sqrt[n]{x},\,x(0)>0,\,T>0$$ by integrating $\int \frac{dx}{\sqrt[n]{x}} = \frac{n}{n-1}x^{\frac{n-1}{n}}$ and considering the integration constant as an ending time $T$, this differential equations could stand the finite-duration solutions (see details here): $$x(t) = \left[\frac{n-1}{n}\left(T-t\right)\right]^{\frac{n}{n-1}}\theta(T-t)\equiv x(0) \left[ \frac{1}{2} \left( 1-\frac{t}{T} + \left| 1-\frac{t}{T} \right| \right)\right]^{\frac{n}{n-1}}$$

With this, I think I could fit the curves to the experimental results by changing the value of $n$, and by the total energy shown on paper 2 (ec. (26)), I will be able to find a figure for the ending time $T$ (hopefully). But after doing it, since maybe it will be a waste of time if the theory don't support the results, I want to see here what happens in the case $n=2$ to analyze if this ansatz make sense.

The core question

Using the finite duration solution $x(t)/x(0)$ at $n=2$, I will use as Ansatz 1 that the radius of the bounded spiral is given by: $$r(t)= a\left(1-\frac{1}{4}\left(1-\frac{t}{T}+\left|1-\frac{t}{T}\right|\right)^2\right)\cong a \cos(\alpha(t))$$

For a disk radius $a=1$ and an ending time $T=100$ (unitless for now), the parametric plot in Wolfram-Aplha of the finite bounded spiral is:

finite duration bounded spiral example

Which look reasonable since the pitch is always decreasing and the spiral achieve a bound in finite time.

Even so, If I plot $\Omega(t)$ using the corresponding $\sin(\alpha(t))$ by trigonometry it will have the form: $$\Omega(t) = \sqrt{\frac{4g}{a\sin(\alpha)}} \equiv \sqrt{\frac{4g}{a\sqrt{1-\left(1-\frac{1}{4}\left(1-\frac{t}{T}+\left|1-\frac{t}{T}\right|\right)^2\right)^2}}}$$ Now, if I plot it using $\sqrt{\frac{4g}{a}}=100$ and $T=7$ then $\Omega(t)$ will have a plot as:

plot from Omega using Ansatz

Which, even when "almost nothing has been done" for fitting the solution to realistic results, if I compared it with the measured values of the wobbling rate plotted on paper 2 they look quite similar:

Image of Figure 5 from paper 2

But even when the result seen reasonable and similar to the experimental values, since Ansatz 1 force Assumption 2 to happen, things related to the reciprocal will necessarily run to infinity because of having a singularity at time $t=T$ (at least here I shown is true for the reciprocal of the solutions of the family of differential equation listed above), and recently I post a question here and here with a similar result, and both were heavily downvoted and then closed, so after continuing I want to know if the Ansatz 1 make sense from the point of view of physicians (this is why I listed every step for finding the ansatz).

Long shot motivation

Not required for the answer if you want to skip it - but interesting

I know that there are things, like photons, which travels billion of km during million years from the stars to arrive to us, so thinking of them like never-ending things is reasonable, so having them been described by a linear differential equation like the Schrödinger equation makes sense, since their solutions are given by power series which cannot model finite duration phenomena as is stated in Fact 4 (actually their probabilities are non-local because at best they vanishes at infinity, this due their modelling through power series).

But also, every time I walk, for me is reasonable to think that there are finite times where each of my feet stop touching the floor, so it looks like there are two kind of objects from the mathematical view of modelling them: things that last forever and things with finite duration, last ones which from Fact 2, to make an accurate model with an autonomous ODE, it should be a Non-Lipschitz differential equation (which I didn't find many in physics).

So thinking in things like the "arrow of time", where are interesting discussions related with that nowadays, the only tool that can figure out its direction is entropy, and it makes me wonder if this is because there are to few people modelling things in physics with finite duration differential equations (so I want to spread the knowledge of their existence).

Think of this, if the Non-Lipschitz point is not enough to figure out from the equations the arrow of time, at least from a finite duration solution it will be easy to figure out where it becomes zero forever after (the point in time where $x(T)=\dot{x}(T)=0$ following the mentioned papers of V. T. Haimo: there the dynamics "die"). Since in forward time it will looks like something that have a discontinuous start at the initial value, that later continuously becomes zero at a time $T < \infty$ (also all its derivatives becomes zero at time $T$ but not necessarily in continuous form), but in backwards time it will look like something that "spontaneously" rise from zero in time $T$ with all its derivatives also valuing zero, it will be violating many conservation rules, indicating that the direction of time is reversed. Also, a never-ending function will be violating causality in backwards time, as is explained for common functions here [video 4].

I think that the singularity in the Euler's Disk is just a mathematical artifact indicating that something in the model have achieved its finite time, but through all the mentioned Facts, I believe that traditional treatments doesn't have properly model them (think in perturbation theory, which is a Taylor expansion, and the quasi-fact 5, they will fail to accurately model points where the solution goes to zero forever, or it has singularities)... maybe using something like a bounded power series like $f(t) = \lim\limits_{M \to \infty}\sum\limits_{k=1}^M a_k(T-t)^k\theta(T-t)$ or even $f(t) = \lim\limits_{M \to \infty}\sum\limits_{k=2}^M a_k(T-t)^{\frac{k}{k-1}}\theta(T-t)$ could made the trick, but I am not sure about this... but what looks certain for me now, is that current models cannot stand solutions of finite duration, so being brilliant in their own explaining behaviors in "almost the whole time", in these critic points they should fail to accurately predict the behavior, probably giving wrong intuitions related to the phenomena they are trying to emulate.... I know is too much work just for an accurate point in time, but think it in this way: How I could properly study the direction of the "arrow of time", If I am don't even able to properly model the time where it stops moving?, at least with a scalar autonomous Lipschitz ODE I will never figure out the ending time, since they can't have finite duration solutions. But since PDEs could have singularities by their own dynamics, and they could be related as the reciprocal of a finite duration solutions (at least for the family of Non-Lipschitz ODEs founded above), I want to know under which conditions a PDEs could stand finite duration solutions, and examples of them (as the Euler's Disk example could be if the assumptions and anzats I made are not wrong).

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Apr 20, 2022 at 21:50
  • $\begingroup$ @ACuriousMind Giving you are an administrator, and also that during the bounty I didn't find a proper answer to the main question, It is possible to you to move this question to the Math.Stack.Exchange? $\endgroup$
    – Joako
    Apr 23, 2022 at 21:57

1 Answer 1


enter image description here

MEASUREMENTS OF THE FINITE-TIME SINGULARITY OF THE EULER DISK R. I. Leine Institute of Mechanical Systems, Department of Mechanical and Process Engineering, ETH Zurich, CH-8092 Zurich, Switzerland, [email protected]

I am not sure if this help you

the linearized equations of motion are:

\begin{align*} & \left[ \begin {array}{c} {\ddot x}\\ {\ddot y} \\ \ddot\varphi \\ \ddot\beta \end {array} \right] = \left[ \begin {array}{c} 2\,{\frac {\sin \left( \Omega\,t \right) \Omega\,\cos \left( \beta \right) \dot\beta \,a}{ \left( \cos \left( \beta \right) \right) ^{2}-6}} \\ -2\,{\frac {\cos \left( \Omega\,t \right) \Omega\,\cos \left( \beta \right) \dot\beta \,a}{ \left( \cos \left( \beta \right) \right) ^{2}-6}}\\ -2\,{\frac {\Omega \,\cos \left( \beta \right) \dot\beta }{ \left( \cos \left( \beta \right) \right) ^{2}-6}}\\ -4\,{\frac {g\sin \left( \beta \right) }{a}}\end {array} \right] \end{align*} and the roll conditions : \begin{align*} &{\dot x}+\sin \left( \Omega\,t \right) \dot\varphi \,a=0\\ &{\dot y}-\cos \left( \Omega\,t \right) \dot\varphi \,a=0 \end{align*}

the simulation results of the non linear equations

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  • $\begingroup$ Thanks for answering @Eli. I saw the paper and is a bit more complicated that the one a cited on the question. They takes less assumptions as I, like keeping the center of mass of the disk in the same vertical point, reason why the plots shows that "petals-like" turns. I found really interesting that in some of their dissipation terms they consider terms that includes the function $\text{sgn}(\cdot)$, which also appear in the non-Lipschitz terms of the paper by V.T. Haimo: $\dot{x}=-\text{sgn}(x)\sqrt{|x|},\,x(0)=1$ as example admits the solution $x(t)=\frac{1}{4}(1-t/2+|1-t/2|)^2$. $\endgroup$
    – Joako
    Apr 22, 2022 at 20:32
  • $\begingroup$ a question: the system of nonlinear differential eqs. between brackets, Is conceptually different from a PDE, right?... or it can be converted into an PDE by some manipulations? $\endgroup$
    – Joako
    Apr 23, 2022 at 6:16
  • $\begingroup$ The only nonlinear state is $\beta$ all others are linearized . I can write you the complete nonlinear equations. Notice that in my equations there is no friction force involved. What is PDE partial differential equations ? Also the simulations results are done with the complete nonlinear equations. $\endgroup$
    – Eli
    Apr 23, 2022 at 6:46
  • 1
    $\begingroup$ The theory is based on ODE second order differential equations that give you nonlinear equations like sin and cos terms . applied Taylor series you obtain the equations that I wrote $\endgroup$
    – Eli
    Apr 23, 2022 at 21:44
  • 1
    $\begingroup$ Thanks. What I mean, thinking in a coin that is wobbling near this ending time, is that center of the coin is not moving in the XY plane, just falling in the Z axis, so instead of the flower-alike patter of your first graph it will look like more like an spiral. I would like to see How your result behave under similar conditions of the analysis I did? (I tried to do it by myself but I get lost with the used geometrical description) $\endgroup$
    – Joako
    May 5, 2022 at 20:36

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