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Is the solution to $\ddot{\theta}+0.021\,\text{sgn}(\dot{\theta})\sqrt{|\dot{\theta}|}+0.02\sin(\theta)=0,\,\,\theta(0)=\frac{\pi}{2},\,\dot{\theta}(0) = 0 \quad\text{(Eq. 1)}$ of finite duration?

I would like to know if the solution is of finite duration, meaning this that the solution reaches zero in finite time and stays there forever after (instead of being "vanishing at infinity"). Looking for a mathematical prove of that (theoretical or numerical, but such as confirms without any doubt if is truly a solution of finite duration).

Numerical solution to Eq . 1

To be explicit about what I am asking for, here is an example: lets think in the equation $\dot{y}=-\sqrt{y},\,y(0)=1$ 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 $\dot{x} = -\text{sgn}(x)\sqrt{|x|},\,x(0)=1$ 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, and demo 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.

I made this question later on a physics forum here, but in the actual question I am only focusing it on the mathematical side of it: please don´t close it as duplicate, or move the other one to the math forum instead


my attempts so far

I am trying to understand the papers by Vardia T. Haimo: Finite Time Differential Equations and Finite Time Controllers but they use tools I don´t have enough background to fully understand (like Liapunov Theory), so I change the classical nonlinear pendulum equation introducing a non-Lipschitz component similar to used on the papers which solutions numerically looks like having finite duration, to have something familiar to work with it.

PS: Obviously, if you give the exact solution $\theta(t)$ it will be awesome!


Added later

Following some comments I have tried to research about Liapunov theory on the book "Nonlinear Dynamics" by H. K. Khalil, and I believe I am understanding the general idea, but is unreal to believe I will master the topic at least before the bounty ends (I am certainly, not a genius).

But in this line, I found the paper Finite-Time Stability of Continuous Autonomous Systems by Sanjay P. Bhat and Dennis S. Bernstein where they used some Liapunov based tools to determine if a function has a Finite-settling-time behavior, but is too advanced to me. Hope someone could answer the question, to have a guide to focus on the specific topic of solution of finite-duration.


another attempt

Following this answer to other question I did, I believe the following:

  • Since Eq. 1 have the zero function as a trivial solution
  • Since the components $\theta$, $\dot{\theta}$, and $\ddot{\theta}$ are all jointly present in Eq . 1.
  • So the Eq. 1 stands the existence of some time $t^*$ such $\theta(t^*)=\dot{\theta}(t^*)=\ddot{\theta}(t^*)=0$, don't meaning here that this/these time $t^*$ exists on the nontrivial solutions, but at least the equation could stand its existence.

To prove that the equation stands a finite-duration solution is equivalent to prove if there exist some "ending time" $T$ where $\theta(T)=\dot{\theta}(T)=0$, because of the following:

If I can choose a finite duration solution $y(t) = \theta(t)H(T-t)$ with $H(t)$ the Heaviside unitary step function: $$H(t) = \begin{cases} 1,\quad t > 0 \\ 0,\quad t \leq 0 \end{cases}$$

I will have that: $$\begin{array}{r c l} \dot{y}(t) & = & \frac{d}{dt}\left(\theta(t)H(T-t) \right)\\ & = & \dot{\theta}(t)H(T-t) + \theta(t)\delta(T-t) \\ & = & \dot{\theta}(t)H(T-t) + \underbrace{\theta(T)\delta(T-t)}_{=\,0\,\text{if}\,\theta(T)\,=\,0} \\ & = & \dot{\theta}(t)H(T-t) \end{array}$$

And equivalently: $$\begin{array}{r c l} \ddot{y}(t) & = & \frac{d}{dt}\left(\dot{\theta}(t)H(T-t) \right)\\ & = & \ddot{\theta}(t)H(T-t) + \dot{\theta}(t)\delta(T-t) \\ & = & \ddot{\theta}(t)H(T-t) + \underbrace{\dot{\theta}(T)\delta(T-t)}_{=\,0\,\text{if}\,\dot{\theta}(T)\,=\,0} \\ & = & \ddot{\theta}(t)H(T-t) \end{array}$$

So, for the second order system with zero as trivial solution, if there exists a ending time $T$ such as $\theta(T)=\dot{\theta}(T)=0$ the finite duration solution could be made by any solution after time $T$ "stiched" with the trivial zero function solution after time $T$ (this because $y(t) \equiv \theta(t),\, t<T$). I don´t know if is easier or harder to prove that exists a point $T$ that fulfills $\theta(T)=\dot{\theta}(T)=0$ than using Liapunov theory tools, but is an idea I have tried, unsuccessfully so far, but maybe it gives an idea of how to solve the problem to anybody else.


Other attempts

  1. Solving Eq. 1 for $\ddot{\theta}(t) = 0$ leads to a $\tan^{-1}$ which never achieves exactly zero. Here is the attempt done by other person in the question of the physics forum. Interestingly are found a time $T \approx 90$ which is similar to the plot ending.
  2. Solving Eq. 1 for $\dot{\theta}(t) = 0$ will lead to the classic equation of the nonlinear pendulum without friction, which solutions are Jacobi Elliptic functions, so since are periodic, there are going to be infinite times where it lands on zero (I have a related question here).
  3. Solving Eq. 1 for $\theta(t) = 0$ leads to $\dot{\theta}(t) = \frac{a^2}{4}(T-t)^2H(T-t)$ as is explained here, which looks promising since it has only one ending time $T$, but with the actual initial conditions it shows to be centered at $T=0$ (at least under my approach), so it don´t work either (since from the plot the only point with $\ddot{\theta}=\dot{\theta}=0$ is at the end, I was hopeful about its validity). Maybe here I am making some mistake, so If someone could try it too and can fix it it will solve the question, since founding $T$ such as $\theta(T)=\dot{\theta}(T)=0$ will answer the problem (if I am not mistaken).

PS (2): 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 clear example, the Euler's Disk toy that show its ending time when the sound stops.


Last update

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:

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. Among the three papers I thought a formal prove to the main question could be done, but I wasn't able to find it by myself.

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    $\begingroup$ This differential equation does not have a closed-form solution in terms of elementary functions. If you want to prove convergence to zero in finite-time, you will have to build a Lyapunov function for that. $\endgroup$
    – KBS
    Mar 24 at 7:55
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    $\begingroup$ Technically, your first example $\overset{.}{y}=-\sqrt y,y(0)=1$ has finite duration because when it hits zero, the derivative remains zero; with the square root defined nonnegative but having a minus sign before it the solution can never rise again. $\endgroup$ Mar 26 at 14:01
  • $\begingroup$ @OscarLanzi That is interesting, I didn't think of it before... I take the example as a slight modification of this video example.. if you know other examples please take a look into this question. $\endgroup$
    – Joako
    Mar 26 at 21:10
  • $\begingroup$ @KBS I tried to to follow your advice and start reading the the book of Khalil, and slowly I am getting the main ideas, but is too advanced for me to trying to use its tools by now: even understanding a bit now, I got lost on the paper of Bhat & Bernstein, since is also required to know metrics, norms, and other advanced math things (at least advanced for the electrician syllabus from where I come). So I want your opinion: Is this question really a TOO HARD thing to prove? (so it is unlikely to get an answer don't matter how much reputation points I would offer on the bounties? $\endgroup$
    – Joako
    Mar 29 at 23:37
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    $\begingroup$ @Joako Yes, your question is difficult as there there is no explicit solution to your equation and it needs to be tackled independently, which may take a lot of time. A local Lyapunov function could be built but this is far from being straightforward. This problem is basically publishable research paper material. $\endgroup$
    – KBS
    Mar 30 at 11:47

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