4
$\begingroup$

The main objective of this question is to figure out if the following differential equation have [finite-duration] solutions: $$ \ddot{\theta}+0.021\,\operatorname{sgn}(\dot{\theta})\sqrt{|\dot{\theta}|} + 0.02\sin(\theta)=0,\quad \theta(0)=\frac{\pi}{2},\,\dot{\theta}(0) = 0 \tag{Eq. 4}$$


Introduction

I am trying to understand the dynamics of the non-linear ODE of the classic pendulum with friction, which is modeled by: $$ \ddot{\theta}+a\,\dot{\theta} + b\sin(\theta)=0,\quad \theta(0)=\frac{\pi}{2},\,\dot{\theta}(0) = 0 \tag{Eq. 1}$$ for some real constants $a$ and $b$ both different from zero.

From the experiments, I believe is natural to expect that the "exact solution" for Eq. 1 should have an "ending time" from where the solution becomes exactly zero forever, this given by the fact that the experimental pendulum does indeed stop moving in reality (so far I know there is no known exact solution for Eq. 1).(For a discussion about the asumption of the pendulum movement stopping in a finite time due pendulum dynamics you can see this question).

But from what I am understanding now from this other question I have made, I am realizing that since Eq. 1 is a Lipschitz ODE, it stands to only have a unique solution. So finite-duration solutions can't be sustained by these kinds of ODEs- even if an analytical solution is obtainable, at best the solutions will lead to functions that "vanish at infinity" instead of having a "true ending time". This is because a function that reaches zero and stays there can't be analytical for the whole real line. As an example, this is what happens with bump functions $\in C_c^\infty$ (I am probably wrong, but it looks like from the Picard-Lindelöf theorem, that its solutions are going to be analytic since the Picard's iterations build a Taylor Series - this is not a fact, but is what I am interpreting of the theorem, so if I am wrong please comment it).

I have found recently a paper Finite Time Differential Equations (V. T. Haimo - 1985) where scalar ODEs with finite-duration solutions are studied. There, the author set all the equations as having an "ending time" at $t=0$, and says: "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." Thus I believe now that to have models that achieve finite-duration solutions, one has to leave the "comfort zone" of Lipschitz ODEs and Uniqueness of solutions (which could be also quite problematic, as the only example I know is the Norton's Dome, which is kind of a fictitious system built over a singularity and indeed there is a lot of discussion related to its solution's definition).

As an example of what I mean with a finite-duration solution, consider this ODE: $$ \frac{y'}{y} = \frac{-2\,t\,(1+|1-t^2|)}{(1-t^2)^2},\quad y(0)=1\tag{Eq. 2} $$ which will have, as a solution $$ y(t) = \frac{(1-t^2+|1-t^2|)}{2} \exp\left[-\frac{t^2}{1-t^2}\right] $$ see plot, which behaves as a smooth-non-analytic bump function $\in C_c^\infty$ within $t=(0,\,1)$ and stays at $0$ after time $t=1$ (there are some definition issues with the differential equation from where the solution becomes zero, and also issues with the solution at $t=1$ but solvable by extending the function through limits keeping its continuity - see details here).

With this idea in mind, if I plot the following equation (this is Eq. 1 with some "arbitrary" values): $$ \ddot{\theta}+0.1\,\dot{\theta} + 0.02\sin(\theta)=0,\quad \theta(0)=\frac{\pi}{2},\,\dot{\theta}(0) = 0 \tag{Eq. 3}$$ which can be seen plotted here, its phase diagram $(\theta,\,\dot{\theta})$ looks like it converges to the point $(0,\,0)$, but because it is a Lipschitz solution, the oscillations will decrease but continue forever in time without arriving at the point $(0,\,0)$ (only achieved at infinity).

Now, using "what is done" in the paper, there are only examples of equations of a specific form: if I introduced a similar Non-Lipschitz component for the friction, "arbitrarily" chosen here just to use it as an example (I don't have any physical explanation for it - if you can give it one it would be awesome), Eq. 3 will look like: $$ \ddot{\theta}+0.021\,\operatorname{sgn}(\dot{\theta})\sqrt{|\dot{\theta}|} + 0.02\sin(\theta)=0,\quad \theta(0)=\frac{\pi}{2},\,\dot{\theta}(0) = 0 \tag{Eq. 4}$$ which can be seen plotted here, where it "looks like" it is indeed behaving similar to Eq. 3, having damped oscillations, but it is also achieving the phase space point $(\theta,\,\dot{\theta}) = (0,\,0)$ in finite time, as my intuition says it should be for a classic mechanic system.


Main Questions

But since these things, being interesting, so far are on-purpose-made and quite speculative, I have the following questions:

  1. Is Eq. 4 actually achieving its ending time in finite time? (a math proof of it)... Here to be explicit for what I am asking for, I have added an example: lets think in the equation $y'=-\sqrt{y},\,y(0)=1$ which has as solution $y(t)=\frac{1}{4}\left(t-2\right)^2$, here the solution is not of finite duration, since after reaching zero at $t=2$, it will start to rising again. But the equation $x' = -\text{sgn}(x)\sqrt{|x|},\,x(0)=1$ will have as solution $x(t) = \frac{1}{4}\left(1-\frac{t}{2}+\left|1-\frac{t}{2}\right|\right)^2$ which indeed is of finite duration reaching zero at $t=2$ and staying there forever (see plot here). I want to know if the solution $\theta(t)$ behaves at the end like the solution $x(t)$, reaching zero at some ending time and staying there forever.
  2. Are there examples of physical models that already use non-Lipschitz representations of friction? For any other kind of model (apart from the Norton's Dome), please move to this more general question.
  3. Is there a more accurate representation of the pendulum with friction, that achieves a finite-duration solution? So far, even though Eq. 1 is widespread as the traditional equation for the "realistic" pendulum, the linear dependence of friction on its derivative is still an approximation, at least from what it is said on Wikipedia page for Drag... so, surely there are other attempts.
  4. Does it make sense for you to work with these finite-duration solutions when modeling classical mechanics? (these question only apply for physicists). Do you ever hear of finite-duration solutions before? Or has Physics forgotten about them?

To be honest, I don´t fully understand what it is said on the paper, but I think that the examples on Wolfram-Alpha will speak for themselves.

PS: About the "arbitrarity" of the selection of the non-lipschitz component, the traditional pendulum model friction as proportional to the rate of change as is explained on Wikipedia as the Newtonian Law of Viscosity, but in the same section is explained that are other models, like the Power Law Model which has components actually similar with which I am using here. I hope that knowing how to model finite duration solution will lead to adapts these classic models to show solutions that indeed behaves as having an ending time $T < \infty$, which I believe is the case on everyday phenomena, as a example, the Euler's Disk toy that show its ending time when the sound stops - I have a related question here).

PS2: An interesting extension of this discussion was made in this other question, hope you can visit to see the details and comment.

PS3: I have improved the model of the drag force into this other question, if you are interested in this question, surely you are going to be checking also this another.

$\endgroup$
5
  • $\begingroup$ "the fact that the experimental pendulum does indeed stop moving in reality" -- what makes you think this is a fact? (a) Do you think you could distinguish a pendulum oscillating with an amplitude less than 1 micron from a pendulum not moving? (b) Don't forget that a real pendulum is connected to other things (such as the Earth) which are moving, and that it has a finite temperature, so there is always some motion in a real pendulum. In fact it is an unrealistic idealization that you can write the motion of a pendulum in isolation from anything else, at the level of precision you want. $\endgroup$
    – Andrew
    Apr 20 at 1:35
  • $\begingroup$ @Andrew Thanks for getting interested. Indeed I can't prove it doesn't stop moving, but since otherwise its movement will disappear immersed in thermal noise, since thermal noise is modeled as Additive White Gaussian Noise (at least in telecom), and Normal distrib. is the maximum entropy distrib. for finite average and power, it means that all information related to the movement is already lost... so there is no data to support is still moving due the pendulum dynamics, so is more simple to think it indeed have stop moving (not meaning is right, but is more accurate from data). $\endgroup$
    – Joako
    Apr 20 at 1:41
  • $\begingroup$ "is more simple to think it indeed have stop moving" -- You say it is more simple, but on the other hand, as you have seen this hypothesis leads to the conclusion that Newton's laws must be wrong for describing the motion of a macroscopic, non-relativistic system, since Newton's laws don't have the solutions you are looking for. I would venture that in fact a simpler hypothesis is that Newton's laws are right, and that what you perceive as a pendulum being still, is actually a pendulum moving with a very small amplitude. $\endgroup$
    – Andrew
    Apr 20 at 1:43
  • $\begingroup$ @Andrew Think for this specific question that is an assumption, but my personal opinion is that the movement due their dynamics have an end, here I am asking it to figure it out. Also, it is not going against Newton's Laws, the proportionality of the friction coefficient to the derivative is an assumption/approximation of a more nonlinear law - it depends on the problem conditions but is accurate enough for common purposes, but as you said, it will never lead to a finite duration, that is why I am trying to look for an alternative. $\endgroup$
    – Joako
    Apr 20 at 1:48
  • $\begingroup$ @Andrew By the way, I am not meaning the proposed example is right, is just a toy model to figure out if this kind of non-Lipschitz component make the solution of finite duration, since is the kind of component used on the paper where I found about this solutions with finite ending times. $\endgroup$
    – Joako
    Apr 20 at 1:51

1 Answer 1

0
$\begingroup$

for this differential equation

$$\ddot \theta+a\sqrt{\dot\theta}+b\,\sin(\theta))=0\tag 1$$

with $~\dot\theta \gt 0$

to obtain the time t for $~\theta(t)\mapsto 0~$ put $~\ddot \theta(t)=0$ and solve for $~\theta(t) ~$

$$-a^2\,\frac{d\theta}{dt}=b^2\sin^2(\theta)\quad\Rightarrow\\ t=-\int \frac{a^2}{b^2\,\sin^2(\theta)}\,d\theta+c =-{\frac {{a}^{2}\cos \left( \theta \right) }{{b}^{2}\sin \left( \theta \right) }}+c $$ $$\theta(t)=-\arctan\left(\frac{a^2}{b^2(c-t)}\right)\tag 2$$ where c is integration constant . $$\theta(0)=\frac{\pi}{2}\quad\Rightarrow\,c=0$$

solving equation (2) for $~t_f~$ with $~\theta(t=t_f)=\epsilon~$ and $~\epsilon~$ is a small value. you obtain

$$t_f={\frac {{a}^{2}}{\tan \left( \epsilon \right) {b}^{2}}}$$

with your data and $~\epsilon=0.01~$ you obtain that $t_f=90~[s]~$


numerical result of

$$\ddot \theta+a\,\text{sgn}(\dot\theta)\,\sqrt{|\dot\theta|}+b\,\sin(\theta))=0\tag A$$ $~a=0.02~,b=0.021~$ enter image description here

$\endgroup$
10
  • $\begingroup$ Thanks you very much for answering, is a really interesting approach, but I don´t fully understand why making $\ddot{\theta}=0$ will do the trick, since it happens also in other places and not only when the solution goes to zero, even so, when decaying to zero $\dot{\theta}<0$ different from the assumptions you use... but even with your approach, Are you able to prove that indeed is being achieved the point $(\theta,\,\dot{\theta}) = (0,\,0)$? (instead of being a solution that "vanish at infinity") $\endgroup$
    – Joako
    Mar 12 at 0:19
  • $\begingroup$ at $t_f$ is $\ddot\theta\approx 0$ Notice that the solution give you the right result, so it can’t be wrong? $\endgroup$
    – Eli
    Mar 12 at 7:57
  • $\begingroup$ I see... and Did you check on your software that after $t=90$ the solution becomes exactly zero forever after? Since the approximation lead to an $\tan^{-1}$ which vanishes at infinity instead of becoming exactly zero, I am not fully convinced if it prove it or not if the solution is indeed achieving the point $(\theta,\,\dot{\theta}) = (0,\,0)$.. hope you can elaborate into that, since is the most important thing I am trying to understand. Thanks you very much. $\endgroup$
    – Joako
    Mar 12 at 12:03
  • 1
    $\begingroup$ with my ansatz is $~\theta(t_f)~=\epsilon~$ the numerical solution $~\theta(90.699)=0.017~$ exact zero you obtain only at t goes to infinity $\endgroup$
    – Eli
    Mar 12 at 14:07
  • 1
    $\begingroup$ As example, here I believe I have found a finite-duration solution to $y' = -\text{sgn}(y)\sqrt{|y|},\,y(0)=1$ using $y(t) = \frac{1}{4}\left(1-\frac{t}{2}+\left|1-\frac{t}{2}\right|\right)^2$, here is not hard to understand that the solution indeed becomes zero after its final time $t_f =2$... since I don´t know the exact solution to Eq. 4, I am trying to see if its solution $\theta(t)$ is also achieving zero in finite time and stays there forever... Hope with this example get clear why I am trying to figure out. $\endgroup$
    – Joako
    Mar 12 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.