0

Understanding Uniqueness of solutions of differential equations - nonlinear ODEs - pendulum example

I am trying to understand If the nonlinear ODE of the classical equation for the pendulum with friction: $$\ddot{\theta}+a\,\dot{\theta}+b\,\sin(\theta) = 0,\quad (\theta_0,\,\dot{\theta_0}) = \left(\frac{\pi}{2},\,0\right)$$ fulfills the conditions of Uniqueness of solutions ($a$ and $b$ are real valued constants, both different from zero).

Arguing on another question, other user tell me that the equations should stand an unique solution, but I believe this user is mistaken but I cannot prove it.

My intuition is the following: Since the equation model a "realistic" pendulum without introducing approximations, and the "physical pendulum" due it stop moving in reality after a period of time, the "exact solutions" of the equation should become exactly zero after an "ending time" and remain zero forever after, so "there is non-uniqueness of solutions through zero in backwards time" as is mentioned in this paper.

I am trying to understand Uniqueness through Wikipedia but my background is not enough (I don´t fully understand the mentioned paper either), so I would like to someone to explain briefly if the equation presented here has or not uniqueness of solutions and why (references to more deep insights are welcome).

Beforehand, thanks you very much.

Mittens
  • 39,145
Joako
  • 1,380
  • To show uniquness let $\theta=x_1$ and $\dot{\theta}=x_2$ then rewrite the equation as the system $$\begin{bmatrix}\dot{x}_1 \ \dot{x}_2\end{bmatrix}=\boldsymbol u\left(\begin{bmatrix}x_1 \ x_2\end{bmatrix}\right)$$ Where $$\boldsymbol u\left(\begin{bmatrix}x_1 \ x_2\end{bmatrix}\right)=\begin{bmatrix}x_2\ -ax_2-b\sin (x_1)\end{bmatrix}$$ Now show $\boldsymbol u$ is Lipschitz continuous, that is, $\exists K>0$ such that for all $\boldsymbol x\neq \boldsymbol x'$, $$\frac{|\boldsymbol u(\boldsymbol x)-\boldsymbol u(\boldsymbol x')|^2}{|\boldsymbol x-\boldsymbol x'|^2}\leq K^2$$ – K.defaoite Mar 07 '22 at 01:55
  • The physical pendulum will never stop, you just did not look closely enough. What will happen is that the amplitude becomes so small that it is no longer completely governed by this equation, it will get forced by random air movements. I think this happens before you need to switch to quantum theory. – Lutz Lehmann Mar 07 '22 at 05:49
  • @LutzLehmann Does it mean instead that the Eq. is not an accurate model? Maybe the friction element must be non-Lipschitz instead of proportional to $\dot{\theta}$... as example, the scalar ODE $y' +\frac{2x(1+|1-x^2|)}{(1-x^2)^2},y = 0,,,y(0)=1$ has as solution $y(x) = e^{-\frac{x^2}{1-x^2}}\frac{(1-x^2+|1-x^2|)}{2}$ which is smooth and at $x=1$ has an "ending time"... this is the kind of solution my intuition says a classic mechanic system should have. – Joako Mar 07 '22 at 21:12
  • @LutzLehmann When studying engineering, in the physiscs laboratory I did an experiment of the classical pendulum with professional measurement equipment (for acquiring samples to see if a Neural Network could model the pendulum dynamics)... and I can tell that the pendulum indeed stops... when it get immersed in noise, it instantly becomes AWGN noise, so you couldn't keep the hypothesis that the pendulum it still moving at any significance level... I believe that the linear damping element is not accurate know, which I am asking here – Joako Mar 09 '22 at 13:29
  • That's what I was saying, at some point the amplitude of the mathematical model falls below the excitations that are caused by random air movements. At that point one could switch to an SDE model, but that is then no longer predictive, it only gives a frame for statistical analysis, something like a Kalman filter to measure the current turbolence of the surrounding air. – Lutz Lehmann Mar 09 '22 at 14:09
  • @LutzLehmann I believe that in that stage the problem lies under what you would "Believe" physics are instead of what you can prove... the SDE model or a Kalmann Filter will be only modeling noise and they will be unavailable to separate it from any remnant movement - since the average is zero and the power is limited, the AWGN distribution is saying that any information available is lost, as a consequence of being the maximum entropy distribution. Since the linear friction relation is indeed an approximation, I think another alternatives will lead to a more accurate solutions. – Joako Mar 09 '22 at 14:18
  • Yes, that is what I am saying. The amplitudes of the exact model becomes insignificant against the noise. The accuracy of the simple mathematical model gets lost as that assumed that the air is a homogeneous medium, even around the interaction with the pendulum. – Lutz Lehmann Mar 09 '22 at 15:04
  • @LutzLehmann I believe I have found a classic example where is kind of obvious the the system indeed has a real-life stop without needing to overthink about the solution becoming immersed under noise, and is the Euler's Disk toy, the noise it makes can confirm that indeed it stops given its wobbling solution has a finite time divergence – Joako Mar 28 '22 at 18:56

2 Answers2

1

It doesn't completely stop, because you just have linear damping. Eventually the amplitude of oscillation is so small that it doesn't really mean anything physical anymore, but mathematically, this model doesn't completely stop. You can see that by analyzing the equation obeyed by the total energy.

Ian
  • 101,645
  • Are you sure?, so far I know there is no known exact solution to that equation, but is a classic example of phase diagrams of differential equations, and from what I have found so far they look that the phase diagrams achieve in finite time the point $(\theta,,\dot{\theta}) = (0,,0)$, instead of being solutions that "vanishes at infinity"... actually my doubt born after read the paper that talks about conditions where ODEs could hold finite duration solutions: they achieve in finite time the point $(\theta,,\dot{\theta}) = (0,,0)$ were the dynamics of the system "dies". – Joako Mar 07 '22 at 01:42
  • @Joako That doesn't sound right for this problem. This equation satisfies the hypotheses of Picard-Lindelof in addition to the possibility of writing down the energy argument. – Ian Mar 07 '22 at 01:44
  • Thanks for answering. I am reading know about the Picard-Lindelof theorem. But is suspicious for me since I think it also stand the conditions of the mentioned paper, but I don´t fully understand them either. Hope you can give it a chance (it should be right also, since is publish on a IEEE journal)... probably I am mistaken, but for me it looks like there is a contradiction. – Joako Mar 07 '22 at 01:50
  • @Joako Maybe there is a tolerance of amplitudes that are considered, or a different damping function. But this one definitely has unique solutions. – Ian Mar 07 '22 at 01:54
  • Do you have a reference for the Total Energy Analysis you mentioned on your first answer? – Joako Mar 07 '22 at 02:32
1

You are mixing contexts with your assumptions and thus get to absurd conclusions.

  • "the equation models a "realistic" pendulum without introducing approximations" is a statement on the combination of physical laws into a mathematical model. The approximations like linearization of the sine for small angles are of mathematical nature.

  • The laws of kinematics themselves are only approximations, even inside physics. They neglect the elasticity of the materials used, turbolence in the air, down to atomic structures.

  • "the "physical pendulum" due it stop moving in reality": Here you switch context to an actual physical pendulum and what can be measured with actual physical instruments. This is not the mathematical model.

If you want to, you can consider the infinitely prolonged oscillations as an artifact of the simplifications in the mathematical model. Every mathematical model is an imperfect reflection of the real reality. Most often the mathematical simulation is too "clean".

On the other hand, the idea of "coming to rest" or "stop moving" is also a fiction, perhaps bringing the apparatus to zero Kelvin might count. Otherwise there will always be small movement, thermal oscillations, air currents, electro-magnetic fields, changes in gravity,... All incredibly small, but still present.

Lutz Lehmann
  • 126,666
  • Thanks for the answer. I totally agree except of the last paragraph, but because know I have a better understanding of what I am trying to ask (the question is indeed kind of out of time). Now I know that there exists finite duration solutions to differential equations, but to be sustained by a differential equation, it have to be non-Lipschitz at least in one point in time. So the issue is more a philosophical question of physics than a mathematical one, and is Why there aren´t used these kind of models that stands these solutions of finite duration for model phenomena that have ending time. – Joako Mar 28 '22 at 19:41
  • ... Mathematically is interesting, because you have now two kind of objects now: things that last forever unaltered (like photons, electrons), and somehow phenomena/things that indeed have finite duration, so their global properties are behaving as local properties, as example, if their differential equation doesn´t have singularities, for a finite duration solution "should" have always a bounded derivative (see my hypothesis here)... so, as example, QM which has power series solutions, can´t stand finite duration solutions. – Joako Mar 28 '22 at 19:47
  • For the pendulum, as an arbitrary example, I modified the dissipative element to see If I can get some finite duration solutions, which I am asking here. – Joako Mar 28 '22 at 19:53