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Examples of Finite-Duration solutions to Autonomous Ordinary Differential Equations ODEs?

Examples of the scalar versions:

  • 1st order: $\dot{x} = F(x)$
  • 2nd order: $\ddot{x} = F(x,\dot{x})$

I have found recently a paper Finite Time Differential Equations (V. T. Haimo - 1985), where its proved that there exists conditions under which first and second order scalar autonomous non-linear ODEs admits Finite-Duration Solutions, meaning here, that they becomes exactly zero due their own dynamics and remains forever on zero after this ending time (they reach zero "from the right", remember that the variable is "time"), so, they are different from thinking on a piecewise section of a solution of a common initial-value problem, neither solutions that are vanishing at infinity, and the same condition of being zero forever after an specific time, make these solutions to fail to fulfill the conditions of Uniqueness, discarding every Linear ODEs of having them (keep in mind this idea: NO LINEAR ODE stands Finite-Duration solutions), and also, since they achieve a point where is becomes the constant zero on a measurable set of points, these solutions cannot be analytical on the whole real line, so power series will have "issues" trying to match them.

But when asking for higher dimensions, as the amazing answer gave here, looks like compact-supported solution in the spatial variables exists (even for Linear Partial Differential Equations - PDEs), but for the time variable the story is totally different, is like there is a "fuzzy zone" trying to define a compact support in time since commonly time is a parametrization in $\mathbb{R}^{n+1}$ dimensions, or is considered as other variable in a space-time configuration, and so far I didn´t found yet any example of a finite-duration-solution to differential equations, not even in the scalar versions.

The closest functions I have found to a Finite-Duration solutions are bump functions $\in C_c^\infty$, which have starts and endings at zero in a closed finite interval in the real line, but unfortunately I have only found one example that fulfill a Delay Differential Equation DDEs $\varphi'(t)=2\varphi(2t+1)-2\varphi(2t-1)$ so I don´t know how to apply the analysis if the cited paper (and also it don't have any known close form), and for other common bump-functions like $e^{-\frac{x^2}{(1-x^2)}},\,|x|\leq 1$ (or its extensions as non-piecewise definitions as are done here, $e^{-\frac{x^2}{(1-x^2)}}(1-x^2+|1-x^2|)$ as example, but the used "trick" introduce issues with the smoothness so maybe they lose their class $C_c^\infty$, and also introduce points where they are not well defined), I have only found non-autonomous nonlinear ODEs, which also don´t fit in the analysis of the paper (thinking here, in using an starting time different from where the bump function starts rising from zero, like $t=0$ at a point in the middle of the "bump").

Hope you can share examples of Finite-Duration Solutions with the autonomous ODE it fulfill (not piecewise sections of ordinary functions like the multiplication by a rectangular function), since they would help me to understand in a more tangible way how they behave.

Motivation/Discussion

Why are so few info on internet about Finite-Duration Solutions? Maybe they are just "too hard" to been researched about? Since I am not a mathematician, I would like to know if I am loosing my time if they are "known" because of being too hard to be solved under any approach.

This is an unknown topic? Actually Wikipedia page for Autonomous Systems don’t talks about the existence of finite-duration solutions in any part (indeed, I recently added myself).

This is specially suspicious, since the description of the second order equations used on the cited paper is quite general, meaning that Uniqueness shouldn't be standing on a wide-spread kind of nonlinear ODEs. Here I would like to note that even if the analysis of the paper have some issues (there are many parts I don't even understand), all the previous and following characterization of these finite duration solutions still stands.

Note that being Autonomous imply that the system is time-invariant, so its law doesn't change with time.

Or Mathematicians/Physicists found it uninteresting?

At least I believe they are of huge importance: in classic mechanics systems, like a pendulum, they stop moving, so their solutions to the equations that describes their dynamics should be of Finite-Duration (so non Unique and also non-analytical), but commonly, for simplicity, friction is not considered (leading to solutions that vanishes at infinity), or they are studied numerically through their phase-space diagrams, where one can check that efectivelly the path dies at some time where $(x',\,x)=(0,\,0)$ (and also more smart and advanced approximations like Perturbation Theory, among others).

If these Finite Duration solutions are as restricted as “continuous and compact-supported functions”, which are always bounded, maybe other similar restrictions will rise for them, and I hope they could give more intuitive explanations to things as the "Least action principle" or the "2nd Law of Thermodynamics", like as a long-shot example, the "Ultraviolet Catastrophe" maybe explained by the Riemann-Lebesgue Lemma, just to give a few ideas...

From now on I will become more speculative, but as example, I have this question where "informally", I think could be argued that for second order systems like which are explained on the cited paper, actually their maximum rate of change will be always bounded, even when these solutions have an unlimited bandwith because they have finite extension in the time variable.

Other line of thought of why they are interesting, is thinking about the simpler form of Schrödinger Equation which is LINEAR (remember first paragraph), so I believe than studying finite-duration solutions could lead to deeply questions at least on physics: there is a huge mathematical difference between phenomena that last forever (like photons) which equations are linear (at least the basic ones, as also is the Maxwell's wave equation), and classic mechanics elements like a pendulum, where at least the exact solution to the nonlinear damped scenario is still unknown, and maybe, because is a solution of these not-so-known finite-duration kind... in this line, as example, it is interesting to see how a bump function as $e^{-\frac{x^2}{(1-x^2)}},\,|x|\leq 1$ behaves near $x=1$ Wolfram-Alpha, where the aggressive exponential behavior at the end (shared by all the bump functions I know so far), makes a lot of sense thinking on the "small-angle approximation" solution of the pendulum, wide known at least in engineering.

I hope you find them as interesting as I do.

My attempts so far

On the mentioned paper, is given as first order ODE example an equation of the form: $$ \dot{x} = -\text{sgn}(x)\sqrt{|x|},\,\,x(0)=1 \tag{A. 1}$$ where Wolfram-Alpha fails to find a solution.

But for example, for the different differential equation: $$ \dot{y} = -\sqrt{y},\,\,y(0)=1 \tag{A. 2}$$ where Wolfram-Alpha is able to find the solution: $$ y(t) = \frac{1}{4}\left(t-2\right)^2 \tag{A. 3}$$

Using this last solution as a reference, I believe that a finite-duration solution to eq. A. 1 is: $$ x(t) = \frac{1}{4}\left(\left|1+\frac{t}{2}\right|+\left|1-\frac{t}{2}\right|-|t|\right)^2 \tag{A. 4}$$ where it can be seen in Wolfram-Alpha that at least for positive time $t>0$ the solutions is indeed solving the equation.I don´t have enough background to formally prove that it is indeed the finite-duration solution, so if you can, I hope you can help me to prove it.

From its plot it can be seen that after time $t=2$ the solution is indeed becoming zero forever, so the triangular function trick is working $x(t) \cong \Lambda^2\left(\frac{t}{2}\right)$, but it also rises the issue on the definition of the differential equation because of its solution being zero forever after the ending time: think of eq. A. 1 in the form $ \frac{x'}{\text{sgn}(x)\sqrt{|x|}} = -1$, at first sight there is no issues, but since it admits solutions of finite-duration, somehow after its final time this problem could be ignored (I hope, please refer to the paper for formality, since I don´t have enough background to explain why it still can be sustained the equation even if their exists a division by zero).

But I still been interesting that the "trick" could be used to find the finite-duration solution (if I am right), since it gives a solution that is not a section of a bump function, as I proposed before and explain here.

Later I have figure out the following: given that the previous result only works for $t>0$, I have tried another alternative solutions: $$ x(t) = \frac{1}{4}\left(\left|1+\frac{t}{2}\right|+\left|1-\frac{t}{2}\right|-|t|\right)^2\,\theta(t) \tag{A. 5}$$ with $\theta(t)$ the standard unitary step function, which is different from zero only on $t\in (0,\,2)$ (plot here), but this time trying to verify if it is solving eq. A.1 becomes a big mess on Wolfram-Alpha, mainly because now there is a discontinuity at $t=0$ where the eq. A.1 is not being matched:

  • $\frac{d}{dt}\left(\frac{1}{4}\left(\left|1+\frac{t}{2}\right|+\left|1-\frac{t}{2}\right|-|t|\right)^2\,\theta(t)\right)=\begin{cases} \frac{t-2}{2},\, 0<t<2 \\ 0,\,t<0\vee t\geq 2 \\ \text{indeterminate},\,\text{otherwise} \end{cases}$ (see Wolfram-Alpha).
  • $\text{sgn}(x)\sqrt{|x|} = \begin{cases} 1-\frac{t}{2},\, 0\leq t \leq 2 \\ 0,\, \text{otherwise}\end{cases}$ (see Wolfram-Alpha).

But, from another happy accident, I found that the following solution indeed solves eq. A.1 in the whole real line: $$x(t) = \frac{1}{4}\left(1-\frac{t}{2}+\left|1-\frac{t}{2}\right|\right)^2\tag{A. 6}$$ which plot can be seen here and it is, at least numerically, fulfilling being a solution of eq. A.1 Wolfram-Alpha, but again, I cannot formally prove it, so any help or comment will be usefull, but at least following what it is said on Wikipedia page for ODEs about Local Existence and Uniqueness, I believe the results are right.

But as the mentioned paper states uniqueness of solution is not granted since equation A. 1 is not a Lipschitz ODE, so I am also wondering if equations A. 4 and A. 6 are actually two different solutions to the Initial Value Problem (IVP) of eq. A. 1: I am not fully aware if a solution of IVP must matched only from time $t_0$ onward or also from the real line that comes before $t_0$, but if is only required onward, non uniqueness could be delivering an extra degree of freedom or flexibility to the solutions, since they could behave as they want after matching the solution between $t_0$ and $t_F$, and I have another question here where I am asking about this, as another example of an IVP, $y(t) = \sqrt{e^{\frac{1-t}{1+t}}-1}$ lives only on the reals in a close interval, and I don´t know if finite duration solution could become complex after the initial time $t_0$.

Joako
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  • Generally this is mostly discussed from the point of view of uniqueness of solution, since this forces non-uniqueness of solution. – Ian Feb 21 '22 at 23:59
  • @ian Indeed, in the paper the author say: "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."... so knowing they are not unique, it will made "easy" or "harder" to find a solutions?? it is somehow less restricted, so I was expecting to find many more examples, and unfortunately, still now I have found zero. – Joako Feb 22 '22 at 00:04
  • What I mean is that this is where you'll find this discussed; for example $y'=|y|^{1/2}$ can become zero in one direction and not the other but not both (unless it is identically zero). – Ian Feb 22 '22 at 00:05
  • @ian Maybe I need to specify more the question, indeed in the paper it is defined what being of finite duration means, discarding things as solution like $\sqrt{1-t}$, that sharply gets to zero, different from "becoming" exactly zero due their own dynamics and remains forever on zero after this ending time (the example becomes complex)... please if you can, take a diagonal read of it, maybe you could suggest a simpler way of state it that the one I have use (also, your example is similar to the one the author gives). On the other hand, Why "math" is so obsessed with uniqueness? – Joako Feb 22 '22 at 00:15
  • @Ian I have added now the difference about solving $y' = \sqrt{y},,y(0)=1$ with solution $y(t)=\frac{1}{4}(t-2)^2$ from solving $x' = -\text{sgn}(x)\sqrt{|x|},,x(0)=1$ wich I believe has solution $x(t) = \frac{1}{4}(1-t/2 +|1-t/2|)^2$.... obviously are the same equations for $t<2$, but I believe the trick is actually solving the differential equation in the whole real line, even when the diff. eq. becomes undefined after $t\geq2$ since it becomes under the analysis of the mentioned paper... but I think is that the difference in finding a finite-duration solution (if I am right) – Joako Mar 12 '22 at 14:31

2 Answers2

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An illustrative example of a system with a finite-duration solution would be Norton’s dome: If you roll a ball with the right speed up this dome, it will come to rest at the top after a finite time (without friction). At this point, the behaviour becomes unclear: The ball staying at the top or rolling down in any direction are admissible solutions in classical mechanics.

However, Norton’s dome is only a thought experiment describing a “point-singular” scenario. Nothing remotely peculiar happens once you change any parameter ever so slightly (under- or overshooting the ball, not aiming straight for the dome’s top, slightly changing the dome’s shape). Thus there are no relevant real parallels. And that’s not even considering friction.

Mind that this different from many other pathologic examples or singularities. While those are usually not reached exactly in reality either or mechanisms beyond a model kick in to prevent them, excessive behaviour already happens in their vicinity. For example if a continuous model is clearly inconsistent at a specific point, we can usually conclude that it will behave unrealistically in the vicinity.

So, unless somebody surprisingly finds a real application of such ODEs, I am not surprised that there is little interest in them. The original paper you cite comes from robotics, in which one might design a system to exhibit such a behaviour. But then even such a system would only approximate the behaviour and one can obtain the desired properties in other ways and without dissecting a mathematical oddity.

Wrzlprmft
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  • Thanks for answering, and sorry for my late response. I didn´t know before the Norton's Dome problem, is quite interesting and unfortunately Wolfram-Alpha don´t solve its equation, but there are a few another related questions in StackExchange where a good insight is given. I understand now from it that non-uniqueness of solutions could be a problem in a problem (even in a quite fabricated one as the Norton's Dome), but differently from your opinion, I believe that given the difficulty Physics have missing and important point when avoiding finite-duration solutions.... – Joako Mar 08 '22 at 15:54
  • As example, thinking in the classic pendulum with friction, the solution to $$\ddot{y}+0.1y'+0.02\sin(y) = 0,,,y(0)=\pi/2,,y'(0)=0$$ shows a infinite duration solution on Wolfram Alpha, but instead using similar to the paper, like $$\ddot{y}+0.021\text{sgn}(y')\sqrt{|y'|}+0.02\sin(y) = 0,,,y(0)=\pi/2,,y'(0)=0$$ give a solution that also shows a damped oscillations, but they achieve a ending point in finite time... since finite-duration solutions are known for being of unlimited bandwidth, studying them will rise restriction as having, as example, bounded maximum rate of change, .... – Joako Mar 08 '22 at 16:00
  • which could help to give more intuitive explanations, for example, to things like the second law of thermodynamics (I hope). – Joako Mar 08 '22 at 16:01
  • I believe that given the difficulty Physics have missing and important point when avoiding finite-duration solutions – I fail to parse that sentence, but: I have never heard of physicist having a serious problem with finite-duration solutions or avoiding them. If you can explain something better using time solutions, great for you, but you’ll have to do that first. – Wrzlprmft Mar 08 '22 at 16:15
  • I know beforehand it is totally an arrogant phrase, but I hope it will be taken with humility since it born from my concern, and not from arrogance: I have recently been aware of their existence of finite duration solutions, and indeed I have been founding results, and since I am a total Mr nobody with only basic knowledge of math (as an electrician), is hard to believe that more smart and educated people have not see them before, since is the only possible reason to me of finding myself "doing new math" that is not on papers or books.... – Joako Mar 12 '22 at 02:07
  • About the importance, think in dynamical systems or second order ODE's (as many examples in mechanics), if the ODE doesn't have singularities on the Right-Hand-Side, I believe that finite-duration solutions, even of being of unlimited bandwidth, they will have always a bounded rate of change, so from the mathematics of the system, their speed will be always limited - this could be important analyzing causality of finite duration phenomena, like their thermodynamics... but surely who will answer that questions doesn't going to be me, I lack the background, and surely I am not that smart. – Joako Mar 12 '22 at 02:13
  • Actually, I am asking that here. – Joako Mar 12 '22 at 02:51
  • I have made some reviews about the Norton's Dome and is quite interesting the following: its solutions or are zero or are the singular solution that suddenly rises from zero and grow as the 4th power law, it is exactly the time-reflected version of a finite-duration solution which comes from some initial value and achieve zero and stays always there, in this case, the singularity works in favor of the mechanical intuition of the problem: as example, the solution I found to the simplest example of the paper decays quadratically (instead of the 4th power of the Norton's equation). – Joako Mar 14 '22 at 18:41
  • I believe now since is of finite duration and has an initial condition, its kind of a boundary value problem so the initial condition "fix" the solution, differently from the Norton's Dome where, since is like a time reflection starting from zero, it doesn't have the border condition that fix the solution leading to its problematic solutions than can rise from zero at any constant value after time zero. – Joako Mar 19 '22 at 22:46
  • I have realized recently that the Norton's Dome equation indeed could be an example but taking different solutions from which are used to analyze the Norton's Dome dilemma: if $\ddot{r} = \sqrt{r}$ indeed is can sustained as solutions the family $r(t) = \frac{1}{144} (T-t)^4\theta(T-t)$ for an ending time $T$ dependent of the initial conditions $r(0) = \frac{T^4}{144}$ and $r'(0) = -\frac{T^3}{36}$... different from the Norton's Dome dilemma, here the solutions are fully determined by the initial conditions since they always have the border condition $r(T)=r'(T)=0$ – Joako Mar 28 '22 at 22:02
  • $\theta(t)$ the Heaviside unitary step function – Joako Mar 28 '22 at 22:08
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So far, following the example given in this answer, it looks like, at least, if a scalar differential equation initial value problem has a Non-Lipschitz point $T>0$ where happens to be true that $x(T)=\dot{x}(T)=0$ and the differential equation stands the trivial zero solution, then is possible to make finite-duration solutions by "stitching" an nontrivial solution for $t<T$ with the zero function for $T\geq t$ (this was noticed in this answer to another question I did).

After asking these questions here and here, I believe that the family of Non-Lipschitz ordinary differential equations for integer $n>1$: $$\dot{x} = -\sqrt[n]{x},\,x(0)>0,\,T>0$$ 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: $$x(t) = \left[\frac{n-1}{n}\left(T-t\right)\right]^{\frac{n}{n-1}}\theta(T-t)$$ with $\theta(t)$ the Heaviside's standard unitary step function. Even so, I am not completely sure but following the style of solutions presented in this paper, I believe the same solutions $x(t)$ will be solutions also for the differential equation: $$\dot{x} = -\text{sgn}(x)\sqrt[n]{x},\,x(0)>0,\,T>0$$ Here I believe there is caution since when $\frac{n}{n-1}$ is even, the order $(T-t)$ or $(t-T)$ doesn't matter on the solutions, but it will be important when taking the derivative so it could indicate that the same solutions answer $\dot{x}=\sqrt[n]{x}$ when it is false.

Also, it could be useful to note that: $$\begin{array}{r c l} x(t) & = & \left[\frac{n-1}{n}\left(T-t\right)\right]^{\frac{n}{n-1}}\theta(T-t) \\ & \equiv & \left[\frac{n-1}{n}\left(T-t\right)\theta(T-t)\right]^{\frac{n}{n-1}} \\ & \equiv & \left[\frac{n-1}{n}T\left(1-\frac{t}{T}\right)\theta(1-\frac{t}{T})\right]^{\frac{n}{n-1}} \\ & = & x(0)\left[\left(1-\frac{t}{T}\right)\theta(1-\frac{t}{T})\right]^{\frac{n}{n-1}} \\ & \equiv & x(0) \left[ \frac{1}{2} \left( 1-\frac{t}{T} + \left| 1-\frac{t}{T} \right| \right)\right]^{\frac{n}{n-1}} \\ & = & \left[ \frac{(n-1)}{2n}T \left( 1-\frac{t}{T} + \left| 1-\frac{t}{T} \right| \right)\right]^{\frac{n}{n-1}} \\ \end{array}$$

added later

Looks like it still works for any real-valued $n>1$ not needing it to be integer.

Joako
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