When working with the Drag Force is typical used on classical mechanics systems the following:

  1. For high speeds it is used the Drag equation which says the drag is proportional to the squared of the velocity: $F = a\dot{x}^2$
  2. For low speeds it is used the Stokes' Law which says the drag is proportional to the the velocity: $F = a\dot{x}$

Then, thinking in the classical nonlinear pendulum with friction on dimensionless units, it equation will be: $$\ddot{x}+F+b\sin(x)=0$$ where traditionally is it used the Stokes' Law approximation. In this specific problem, the only component which can be changed is the drag force since all other terms are determined by geometry and the classic assumptions of a stiff weightless rod, and the only friction force present is air drag.

Recently in this question I learned that the only possible way for a differential equation $\ddot{x}=G(\dot{x},x)$ to admit a solution that achieve a finite extinction time if by having at least one point where the differential equation $G(\cdot,\cdot)$ is Non-Lipschitz (references on the mentioned question).

I would like to find a model for the pendulum that stops moving in a finite time. But since both mentioned alternatives will made $G(\cdot,\cdot)$ locally-Lipschitz, I believe neither of them are able of having a finite ending time.

This is why I would like to know if is possible to have a Drag Force that is Non-Lipschitz under current fluid mechanics theories, hopefully with examples.

In the past I made a related question buy just guessing an ansatz $F=a\,\text{sgn}(\dot{x})\sqrt{|\dot{x}|}$ but I was unable to prove if it really achieving a finite extinction time.

Added later

I have been thinking in my previous ansatz and in the two common models for the Drag Force, and I realize the following:

I haven't found many locally-non-Lipschitz (LNL) functions apart of $\text{sgn}(x)|x|^\alpha,\,0<\alpha<1$ "friendly enough" (as example, $x\log(x^2)$ is also LNL but has sign changes for $x>0$), so for now I will following using the form $\text{sgn}(x)\sqrt{|x|}$.

With this, if I "make" a drag force of the form $F_*=a\,\text{sgn}(\dot{x})\sqrt{|\dot{x}|}\left(1+|\dot{x}|^{\frac{3}{2}}\right)$ it will be fulfilling:

  1. from speeds $0.05<\dot{x}<1$ the force $F_* \approx a_1\dot{x}$ relatively well, so it behaves similar to Stokes' law at law speeds - at least it "look like" is linear/affine function, since a little offset is required for made the match.
  2. from larger $\dot{x}$, actually it behaves really closed to a quadratic function $F_* \approx a_2(\dot{x})^2$, so it could be modelling the drag force for high speeds
  3. from really small speeds $\dot{x}<0.05$ the force $F_*$ keeps the LNL behavior so keeps open the possibility if achieving finite extinction time.

Here the plots of the three regimens to view exactly what I have been comparing: Comparison of Stokes'law, drag force, and proposed term

Since for the linear Stokes' part and the quadratic Drag force there are already physical justifications (are widely used), I would like to know if the sudden increase at the beginning made sense from the real behavior of fluids (since right know is just a mathematical trick oriented to achieve a finite extinction time).

As example, if I used Wolfram-Alpha to simulate the nonlinear pendulum with friction under the effect of this proposed drag force $F_*$, using arbitrary constants just for having useful plots on the free version of Wolfram-Alpha, I will have the following equation been studied: $$\ddot{x}+0.021\cdot\text{sgn}(\dot{x})\sqrt{|\dot{x}|}\left(1+|\dot{x}|^{\frac{3}{2}}\right)+0.2\cdot\sin(x)=0$$

Which under the initial conditions $x(0)=\frac{\pi}{2},\,x'(0)=0$ shows a decaying oscillating behavior as expected for a pendulum, but unfortunately the graphs don't shows the full extension in time:

modified pendulum for x(0)=pi/2

So, for trying to achieve a full phase space diagram, I started the simulation with a lower angle for the pendulum: $x(0)=\frac{\pi}{25},\,x'(0)=0$ and it looks like indeed under this configuration the pendulum is achieving a finite extinction time, but unfortunately the plot versus time is really bad, but it looks it have already stops moving:

modified pendulum for x(0)=pi/25

This is interesting I think, since every traditional model for the pendulum I am aware of, will be continually spiraling towards the point $(0,0)$ forever without never achieving it.

I know beforehand that the hypothesis of this pendulum equation been achieving a finite extinction time must be mathematically proven first (which could be particularly hard since the differential equation is highly non-linear), and because of this, I want to focus the question in physics aspects of considering a Drag Force with a Local-Non-Lipschitz behavior:

For people that works with fluid mechanics,Does makes sense physically speaking, considering a Drag Force modeled as $F_*$? There are any experimental examples similar, or that could support this kind of hypothesis of having a local-non-Lipschitz behavior?

I know that not every physical phenomena needs to achieve a finite ending time, this is not what I am claiming, but I believe there are a lot of phenomena that indeed does that, and I want a way to accurate model them.

2nd Added later - focus for the bounty

Note please that for very low speeds the proposed drag force $F_*=a\,\text{sgn}(\dot{x})\sqrt{|\dot{x}|}\left(1+|\dot{x}|^{\frac{3}{2}}\right)$ rises with infinite slope from zero (this due the same term it made it locally non-Lipschitz, requirement for having a finite stopping time). I search for Drag Force vs air speed tables for a spherical profile but the one I found start from speeds higher than 10 km/h, so I didn't get an opinion in favor or against it: when experiments left the "random thermal zone", Does Drag Force rise as the squared root? Or is already linear as the Stokes' Law? Or is quadratic? Or another?

I imaging this "weird situation" as an analog of a perfect elastic collision of a ball with a wall, let say as example, its position is described in one dimension as $x(t)=|t|$ so it hits the wall at $t=0$. Then, its speed profile is going to be discontinuous but bounded $\dot{x}(t)=\text{sgn}(t)$ so $\|\dot{x}\|_\infty =1 << c$, so nothing is "weird" yet. But the sudden change in direction leads to a singularity in the aceleration profile $\ddot{x}(t)=2\delta(t)$ so $\|\ddot{x}\|_\infty \to \infty$. This is why I believe that the behavior of $F_*$ at speed $\dot{x}\to 0$ is not violating any physics law... but I am not really sure (maybe my previous example is ill-defined)... but since currently are being studied finite time singularities in the Navier Stokes equation, I don't believe this is going an issue for the model.

3rd Added later - similar papers

I have just found the following papers under the term sublinear damping that shows example in physics of equations really similar to the proposed example:

There are mention similar equations in physics like the Coulomb damping where similar issues like having solutions of finite ending times are shown, with their consequences of non-uniqueness of solutions' issues and the lost of the differential equation time-symmetry.

Particularly, the first one, since graphically the solutions shows having a decaying envelope, I think could be proving than when the solutions enters the small-angle approximation regimen $\sin(x) \approx x$ it will be indeed being a solution with a finite extinction time.

I would like to explicitly note why I am not using the standard ansatz for the drag force $F_{\text{drag}}\propto (\dot{x})^2$ since as is shown here for the equation $$\ddot{x}+0.021(\dot{x})^2+0.2\sin(x)=0, x(0)=\frac{\pi}{2}, \dot{x}(0)=0$$ their solution aren't showing the expected decay one can see on the experimental pendulums. This is commonly solve using an ansatz for the drag force $F_{\text{drag}}\propto \dot{x}|\dot{x}|$, which as can be seen here for the equation $$\ddot{x}+0.021\dot{x}|\dot{x}|+0.2\sin(x)=0, x(0)=\frac{\pi}{2}, \dot{x}(0)=0$$ their solution are indeed having the decaying behavior.

But, since the function $f(x)=x|x|$ is locally Lipschitz (see here), it cannot be used for having a singular point on a finite time where uniqueness could be broken, so this ansatz wouldn't have solutions of finite duration, this is why I chose to work with $F_*$.

By the way, please keep in mind these examples, since for having a decaying behavior it was required to choose a differential equation that is not fulfilling the time-reversal-symmetry , and I believe no differential equation that have "finite extinction time" will stand it either.


This was explained on the mentioned question aboved, and I first left it out to keep this question short, but since answers and comments are becoming focused on this I will post it also here

I am trying to find a mathematical framework to work with dynamical systems under the assumption: the system achieves an end in finite time, which seems really logical to me through daily life experience, but I have found this is really messy when it is translated to math.

Current physics solutions commonly at best "vanishes at infinity", which is not exact but fairly good for almost every possible application, so good, that I have had discussions with many people that affirm as true that movement last forever due the system dynamics, which at least under the assumption that thermal noise is Gaussian, you can say if it still moving (neither if it has stopped) after movement get inmersed in noise, since already all the information is lost: Gaussian distribution is the Maximum entropy probability distribution for a phenomena with finite mean and finite power. Don't meaning the assumption of never ending movement is wrong, but affirming is true is indeed false. Even so, If thermal noise affect "linearly" the system solution $x(t)=x_h(t)+x_p(t)$ as an external random forcing, it will, in principle, affect only the particular solution $x_p(t)$, so it could be considered the solution of the homogeneous equation $x_h(t)$ as finite duration solution without interfering with the behavior under the noisy regime (for nonlinear systems it could not be fulfilled in this way).

This subtle distinction is seen for example, in the Galton board toy, where many people affirm that somehow order have risen from chaos, due the appearance of the Normal Distribution, when the right interpretation is that all the information of the path each ball have take when falling is already lost, and the Gaussian distribution is the maximum possible disordered way of been for this system (to have an intuition of why there is a lobe in the middle, take a look to Concentration inequality).

On the original question motivation it reviewed and extended some possible consequences of working with models that achieves finite extinction times (like breaking time-reversal symmetry, keeping causality, among others).

But for the specific case of a classical system as the pendulum, the regimen where random noise takes over will never been modeled through a classic ODE, but instead stochastic calculus is going to be required like Ito's diffussions and brownian motion, which is far outside modelling classical things with or without finite extinction times.

So I am focus in the system model under the classic assumptions. As example, the dimentionless equation for the Norton's Dome: $$\ddot{r}=\sqrt{r}$$ Under initial conditions $r(0)=\frac{T^4}{144},\,T>0$, and using $\theta(t)$ as the Heaviside step function, could admit finite extinction solution: $$r(t)=\frac{1}{144}(T-t)^4\theta(T-t)$$ so having finite extinction time for a classic 2nd order ODE is achievable without get into the discussions of molecular noise... is the dynamics of the macro-model which is into scrutiny, and I am exploring this alternative of finite duration solutions.

Being true are a lot harder to manage (they require abandon uniqueness which is huge - but at least from the already published examples of finite-extinction-time solutions listed here, uniqueness is hold within the time domain before its goes extinct), I think they could improve some existent models: as a long-shoot example, the same finite duration solution for the norton's dome mentioned above, if I take its reciprocal $z(t)=\frac{1}{r(t)}$, I will get the following diff. eqn.: $\ddot{z}+z\sqrt{z}-2\frac{\dot{z}^2}{z}=0$ which can be solved by $z(t)=\frac{144}{(T-t)^4}$, so at least for some finite-time blow-up behaviors, they could happen due some other variable on their denominator is indeed achieving a finite ending time, which I am asking for the Euler's Disk Toy example here (which show it have stop moving with the abrupt interruption of the wobble sound)... which if even is not the case on this example, at least this make sense for being taking in mind when doing analysis.

Unfortunately I don't have access to a lab for doing experiments by myself and check if the solutions could be fit accurately to experimental results, but at least numerically, in online Octave I check for small initial angles with arbitrary constants, and it looks is indeed having a finite extinction time (I cannot formally prove it so far).

f = @(t,y) [y(2);-0.1*sign(y(2))*sqrt(abs(y(2)))*(1+(abs(y(2)))^(3/2))-9.8/0.5*sin(y(1))]; 
t0 = 0; y0 = [pi/3600;0];
[ts,ys] = ode45(f,[t0,5],y0,opt);

Last update

I found later this paper:

where the autors, in equation $(19)$ use a similar drag force of the form $F_d = b\ |\dot{y}| + c\ (\dot{y})^2$, so at least a description through two components doesn't look like a complete insane approach. Also on equations $(59)$ and $(60)$ they introduce a Drag Force $F_d = b\ \dot{y} + c\ \dot{y}|\dot{y}|$ for a more accurate description of the air effects on the pendulum.

But when comparing the model $F_d = b\ \dot{y} + c\ \dot{y}|\dot{y}|$ with the model I am using above, I found that the later made a better fit for a longer interval of speeds: a Drag Force of the form: $$F_d (v) = a\ \text{sgn}(v)\sqrt{|v|}\left(\frac{1}{q}+|v|^{3/2}\right)$$ fits "good" a linear profile "$a\cdot v$" if $2 \leq q \leq 3$ for low speed ($0.1 < v < \frac{1}{2}$), where I arbitrarily choose $q=2\sqrt{2}$ so $F_d\left(v=\frac{1}{2}\right) = a\cdot \frac{1}{2}$, and also, by good luck, it also fits better the quadratic behavior of the classic Drag Force without loosing the behavior near zero that should allow the existence of solutions of finite duration.

As example the images compare the following drag forces for a common constant $a=1$: $F_d = a\ v$, $F_d = a\ v^2$, $F_d = a\ \left(\frac{v}{2}+v\ |v|\right)$, and the "improved" version: $$F_d = a\ \text{sgn}(v)\sqrt{|v|}\left(\frac{\sqrt{2}}{4}+|v|^{3/2}\right)$$

Ultimate drag force comparisson

In principle, I believe that the constant $a$ should be $a = \frac{1}{2}\rho\;\!C_D A$, but since the Drag Coefficient is also speed-dependent rigorously speaking $C_D(v)$, I am not really sure if this modified Drag Force will fit by itself the Stokes' Law at low speeds, and obviously experimental analysis is required to see if this mathematical construction is fitting properly reality.

But at least for the pendulum examples reviews above, it still looks is behaving as a decaying pendulum: $$\ddot{x}+0.021\cdot\text{sgn}(\dot{x})\sqrt{|\dot{x}|}\left(\frac{\sqrt{2}}{4}+|\dot{x}|^{\frac{3}{2}}\right)+0.2\cdot\sin(x)=0$$

Modified drag force on pendulum

By playing in online Octave with the updated Drag Force I wasn't able to determine if there is a finite time $T$ where the solution becomes exactly zero, but at least it shows that indeed exists a finite time $T$ where it stops oscillating, acquiring only a decaying behavior (differently from the classic solutions for the nonlinear pendulum with friction under the small-angle approximation, that oscillates forever as decaying for vanishing at infinity).

f = @(t,y) [y(2);-0.1*sign(y(2))*sqrt(abs(y(2)))*(sqrt(2)/4+(abs(y(2)))^(3/2))-9.8/0.5*sin(y(1))]; 
t0 = 0; y0 = [0.0001;0];
[ts,ys] = ode45(f,[t0,5],y0,opt);

end of oscillations 1

f = @(t,y) [y(2);-0.1*sign(y(2))*sqrt(abs(y(2)))*(sqrt(2)/4+(abs(y(2)))^(3/2))-9.8/0.5*sin(y(1))]; 
t0 = 0; y0 = [0.00001;0];
[ts,ys] = ode45(f,[t0,10],y0,opt);

end of oscillations 2

In the following picture it can be seen the overall behavior starting from the angle $\frac{\pi}{2}$:

f = @(t,y) [y(2);-0.1*sign(y(2))*sqrt(abs(y(2)))*(sqrt(2)/4+(abs(y(2)))^(3/2))-9.8/0.5*sin(y(1))]; 
t0 = 0; y0 = [pi/2;0];
[ts,ys] = ode45(f,[t0,100],y0,opt);

overall behavior

  • 3
    $\begingroup$ I'm not a fluid dynamics expert, but I think the emphasis is misguided... in real life, masses don't actually come to perfect rest within a finite time. Instead, their original motion just damps until you can't see it with your detector. Or, if you have a really good detector, eventually its motion will be some low-amplitude wobbling due to seismic vibrations, thermal effects, etc. It seems you're trying to impose too much mathematical idealization on a problem which is inherently messier than that. $\endgroup$
    – knzhou
    Jun 10, 2022 at 20:47
  • 1
    $\begingroup$ I dropped your proposed equation into Mathematica. Around $t = 120$, it transitions from oscillating into a decay-like behavior that (by eye) looks like it obeys $x \propto t^{-1}$. This would be consistent with the $\sqrt{|\dot{x}|}$ and $\sin(x)$ terms being dominant in the equation of motion, since (in the limit of small $x$) both would be proportional to $1/t$, with $\ddot{x} \propto t^{-3}$ being negligible. $\endgroup$ Jun 14, 2022 at 19:50
  • 1
    $\begingroup$ The paper you link does say that there are two orbits that reach the rest point in finite time. However, they also argue that "most" orbits take infinite time to reach the rest point. This would imply finding these orbits via numerical techniques would be difficult (since rounding & step-size errors would tend to move you off of these "special" trajectories.) $\endgroup$ Jun 15, 2022 at 14:18
  • 2
    $\begingroup$ Hi, we've noticed that you have made a large number of minor edits to this post. Please be mindful that every edit bumps the post in the "active" tab of the site and try to make your edits substantial. If you foresee improving this post repeatedly, maybe collect several edits and make them in one go instead of submitting them individually. $\endgroup$
    – Chris
    Jun 17, 2022 at 19:38
  • 1
    $\begingroup$ In case @Chris' comment above was not clear enough: It is not acceptable to edit your post (the tags or the body) in minor ways solely because you are "looking for answers" as your edit description says. Edits should improve the posts they are changing and nothing else. Please stop performing edits like that, otherwise your posts might be locked against edits entirely or you might find yourself suspended. $\endgroup$
    – ACuriousMind
    Jul 2, 2022 at 17:51

2 Answers 2


It may be mathematically possible to have a model with an exact finite stopping time by choosing a non-Lipschitz function in the differential equation. But it is nothing more than a mathematical game, flexing your muscles. Physical meaning I see none whatsoever.

What you are asking about is a mathematical pathology of the drag force at vanishing velocity. Now you need to remind yourself that every physical model has a certain range of validity that can either be explained by how the model is built or tested in experiment. And this is not particular to fluid dynamics or the example you gave, but can be found across all disciplines of physics. Let's study your example.

Can we test whether your force model is valid? In approximation, yes, but not to the last detail (say differences of $10^{-100}\,$m/s :P) when approaching $\dot{x} = 0$ - which is the part we need for the non-Lipschitz point. Once you can't measure the velocity difference from zero you might as well say the object has stopped.

Let's also look at the model validity from a more conceptual point of view: We have a macroscopic body embedded in a fluid that is modeled as a continuum. Influences on the body are described as macroscopic pressures or friction forces. If you try to use the model to describe velocity differences that are so small that one molecule bumping into your macroscopic object can be their cause, the whole model goes out the window, obviously.

  • $\begingroup$ Thanks you very much to get interested. I have added now a section was on other question in the aim of focus the discussion in the classic macroscopic modeled where the question is framed. $\endgroup$
    – Joako
    Jun 10, 2022 at 21:55
  • $\begingroup$ Te be precise, answers like yours are interesting but too general: says nothing about the ansatz itself more that could be wathever thing. As example, Is the square root at the beginning violating any physics law? (I don't think so since there are currently studies of singulatities in the Navier Stokes eq.)... also, Drag Force's tables from air starts linearly when they got out from the noisy zone of movement? Quadratically? Or something that resembles the square root behaviour? Or others? $\endgroup$
    – Joako
    Jun 10, 2022 at 22:41
  • $\begingroup$ I have search for these tables on a spherical profile in engineering books but they started on speeds higher than 10 km/h... and in google there are a lots of graphs with mixed results or using things I don't understand. This is why I am aimming for a specific answer. Thanks anyway for taking the time to answer. $\endgroup$
    – Joako
    Jun 10, 2022 at 22:50
  • 1
    $\begingroup$ I think I was rather specific: The model is meaningless as a description of the physical phenomenon once the velocities are too small. Because we are unable to make measurements to test the model and violate assumptions that are used to build the model. This is more than "whatever thing" about the ansatz, I would say. $\endgroup$
    – kricheli
    Jun 11, 2022 at 4:00
  • $\begingroup$ Compare it to another singularity: A planet's gravitational potential is in a wide range of lengths described by $\phi \sim \frac{1}{r}$. But this description has limits. When we are studying what happens right at the singularity we are violating an assumption that was used to built that model: that the mass is point-like. $\endgroup$
    – kricheli
    Jun 11, 2022 at 4:04

Let's look at what you are after from a half physical and half mathematical point of view.

Assume that you have a drag force which satisfies the following sensible conditions:

1- It is local in time, meaning that the value of the force at any instance depends only on the state of the system in the same moment.

2- It only depends on particle's velocity, not its position, acceleration or higher(>2) order derivatives of its position.

3- It changes sign if we take $x\to -x$ or $t\to -t$: $$ F_{drag}=F(\dot{x})=-F(-\dot{x})$$

Let's further assume that under the presence of this drag, a particle stops after a finite amount of time we call $t=t_{f}$.

Now consider the system at any time after $t_f$ that we call $t_b$, the equation that describes the motion is: $$m\ddot{x}+F(\dot{x})+k\sin{x}=0$$

If we turn the time backward $t\to -t$ we get a time reversed equation: $$m\ddot{x}-F(\dot{x})+k\sin{x}=0$$

If we solve this equation using initial conditions $\dot{x}(t_b)=0$,$x(t_b)=0$ we clearly will get $x(t)=0$ for later times. But, at the same time, time reversing the equation of motion is like playing a video backward, so what we expect is to see the particle moving backward in time toward its initial position. So we face a contradiction here and therefore such force cannot exist.

Of course, for a linear or quadratic drag force, the position never becomes exactly equal to zero, and we always can time reverse the system to reach the initial state.

  • $\begingroup$ thanks you very much for taking the time for answering. I don't fully get the answer, but I think that $F_*$ behaves as Stokes's Law in what it concern to their signs under the transformation $t \to -t$ (equivalently I think to $\dot{x}\to -\dot{x}$). Even the Drag Force (which isn't), is many times used as $\propto |\dot{x}|\dot{x}$ instead of $\propto \dot{x}^2$ at least for what I found in Google due it made some issues: as example,... $\endgroup$
    – Joako
    Jun 14, 2022 at 22:28
  • $\begingroup$ ... if I plot in Wolfram-Alpha this eq.: $x''+0.021*(x')^2+0.2*\sin(x)=0, x(0)=\pi/2, x'(0)=0$ it solutions don't show the attenuation it should have, and if instead I plot $x''+0.021*x'*|x'|+0.2*\sin(x)=0, x(0)=\pi/2, x'(0)=0$ the attenuation behavior appears as it should (this, from experimental experience). $\endgroup$
    – Joako
    Jun 14, 2022 at 22:31
  • $\begingroup$ But from the main point you mention, I think is indeed the lost of this time-symmetry behavior what it makes them interesting (at least one of many), since it is always said that in physics the only equation that show the direction of flow of time is entropy, and it doesn't made sense I think in a system is achieving a finite time end - time symmetry maybe is not always an desired property in physics, and nowadays is always there because the considered equations are traditionally Lipschitz diff. eqns. always holding uniqueness of solutions (an commonly, also non-piecewise analytical fns). $\endgroup$
    – Joako
    Jun 14, 2022 at 22:38
  • 1
    $\begingroup$ Don't meaning here is always right, but is why I am trying to explore. Also, going from zero in backward times is a no-so-easy topic there since uniqueness is broken, I think is equivalent to going in normal time on the case of the Norton's Dome example, I also believe it should be zero forever, but there is an example where many informed people are arguing that assumption. $\endgroup$
    – Joako
    Jun 14, 2022 at 22:42
  • $\begingroup$ I have added a "3rd added later" section where some papers with really similar differential equations are used as models in physics, having finite ending time with all the consequences it carries, like having differential equation that don't holds being symmetric in the time variable change $t\to -t$. $\endgroup$
    – Joako
    Jun 15, 2022 at 4:10

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