Are these equations properly defined differential equations?

Modifications were made to a deleted question to re-focus it. I am trying to find out if there exists any exact/accurate/non-approximated mathematical framework to work with physics under the assumption that phenomena have finite ending-times, instead of solutions that at best vanishes at infinity.


Definition 1 - Solutions of finite-duration: the solution $y(t)$ becomes exactly zero at a finite time $T<\infty$ by its own dynamics and stays there forever after $(t\geq T\Rightarrow y(t)=0)$. So, they are different of just a piecewise section made by any arbitrarily function multiplied by rectangular function: it must solve the differential equation in the whole domain. (Here I just pick the shorter name its look more natural to me for several I found: "finite-time", "finite-time-convergence", "finite-duration", "time-limited", "compact-supported time", "finite ending time", "singular solutions", "finite extinction time", among others).

Recently, in this answer I figure out that the following autonomous scalar ODE: $$\dot{y}=-\sqrt[n]{y} \tag{Eq. 1}\label{Eq. 1}$$ Could stand the finite-duration solutions, for any real-valued $n>1$ and positive ending time $T>0$: $$ y(t)=\left[\frac{(n-1)}{n}\left(T-t\right)\right]^{\frac{n}{(n-1)}}\,\theta(T-t)\equiv y(0)\left[\frac{1}{2}\left(1-\frac{t}{T}+\left|1-\frac{t}{T}\right|\right)\right]^{\frac{n}{(n-1)}} \tag{Eq. 2}\label{Eq. 2}$$ with $\theta(t)$ the standard Heaviside step function.

But later, working with the solution of \eqref{Eq. 2} I figure out that if I take its derivative I will end with: $$\begin{array}{r c l} \dot{y} & = & -\left[\frac{(n-1)}{n}\left(T-t\right)\right]^{\frac{1}{(n-1)}}\,\theta(T-t)+\underbrace{\left[\frac{(n-1)}{n}\left(T-t\right)\right]^{\frac{n}{(n-1)}}\,\delta(T-t)}_{=\,0\,\text{due}\,(x-a)^n\delta(x-a)=0} \\ & = & -\left[\frac{(n-1)}{n}\left(T-t\right)\right]^{\frac{1}{(n-1)}}\,\theta(T-t) \\ & = & -\frac{n}{(n-1)}\frac{1}{(T-t)}y(t) \end{array}$$ Which implies that the solutions $y(t)$ are also solving the non-autononous scalar ODE (is time-variant since the time $t$ is explicitely involved): $$\dot{y}+\frac{n}{(n-1)}\cdot \frac{y}{(T-t)} = 0 \tag{Eq. 3}\label{Eq. 3}$$

This last equation of \eqref{Eq. 3} resembles very much this another question were I found that the finite duration solution: $$z(t)=\exp\left(\frac{t}{(t-1)}\right)\cdot\theta(1-t) \tag{Eq. 4}\label{Eq. 4}$$ can solve the equation: $$\dot{z}+\frac{z}{(1-t)^2} = 0,\quad z(0)=1\tag{Eq. 5}\label{Eq. 5}$$

So I started to try another finite-duration solutions to see if their differential equations follows the same "trend" of having terms $(T-t)$ on it: which means that an standard integration constant is becoming part of the differential equation as its finite ending time $T$.

Main analysis

On this question is found that similarly for equations \eqref{Eq. 3}: $$x(t)=\frac{1}{4}(2-t)^2\theta(2-t) \tag{Eq. 13}\label{Eq. 13}$$ is a finite duration solution to: $$\dot{x}=-\text{sgn}(x)\sqrt{|x|},\,x(0)=1\tag{Eq. 14}\label{Eq. 14}$$ so I will modify this solution to see what happens: $$q(t) = \frac{1}{4}(2-t)^2\cos(20\,t)\,\theta(2-t)\tag{Eq. 6}\label{Eq. 6}$$ which can be verified is solves the equation: $$\ddot{q}+\frac{4}{(2-t)}\cdot \dot{q}+\left(400+\frac{6}{(2-t)^2}\right)\cdot q = 0,\quad q(0)=1,\,\,\dot{q}(0)=-1\tag{Eq. 7}\label{Eq. 7}$$

Again, on \eqref{Eq. 7} there terms $(2-t) \cong (T-t)$ indicating that there exists solutions that are behaving like finite duration solutions (see plot of \eqref{Eq. 6}).

But these "construction" of making finite duration solutions where these two conditions are met:

  1. The differential equation support the zero solution, and
  2. The differential equation also has at least one singular point in time $t=T$ where $x(T)=\dot{x}(T)=0$

both mentioned on this answer, becomes less obvious where more than one variable is considered. Let think and a slight modification to the function of this answer:

$$U(t,x) = \exp\left(1-\frac{1}{(1-(x-t)^2)}\right)\theta(1-(x-t)^2) \tag{Eq. 8}\label{Eq. 8}$$

it will show to be compact supported in space as it moves in the time variable, fulfilling the standard wave equation with a velocity $c=1$: $$\left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\right)U(t,x)=0 \overset{c=1}{\Rightarrow} U_{tt}-U_{xx}=0 \tag{Eq. 9}\label{Eq. 9}$$

Here the partial equation doesn't show the term $(T-t)$, but it does make sense if the countour plot of \eqref{Eq. 6} is viewed:

Eq 8 plots

It is only compacted-supported in space and it is still never-ending in time.

But if I test this modified version of \eqref{Eq. 8} mixed with the terms of \eqref{Eq. 2}:

$$E(t,x) = U(t,x)\cdot \frac{1}{4}(2-t)^2\,\theta(2-t) = \exp\left(1-\frac{1}{(1-(x-t)^2)}\right)\theta(1-(x-t)^2)\cdot\frac{1}{4}(2-t)^2\,\theta(2-t) \tag{Eq. 10}\label{Eq. 10}$$

where this time it can be seen in its plot that is indeed behaving as also a finite duration solution in time:

Eq 10 plots

It could be verified that is solves the equation: $$E_{tt}-E_{xx}+\frac{4}{(2-t)}E_{t}+\frac{6}{(2-t)^2}E = 0 \tag{Eq. 11}\label{Eq. 11}$$ within the domain \eqref{Eq. 10} is different from zero (see here). So again the terms $(T-t)$ are required for having finite-duration solutions.

I beforehand know that some of these functions should be defined piecewise because has a undefined value, as example: $$ f(t) = \begin{cases} \exp\left(\frac{t}{(t-1)}\right),\quad t < 1 \\ 0, \qquad t\geq 1 \end{cases} \tag{Eq. 12}\label{Eq. 12}$$ instead of the one-line version of \eqref{Eq. 4}, so I ask you now to forgive this shortcut for now: the functions still continuous on these points and for me are easy to work with them finding their differential equations on this way (I have a extended discussion of this here), but if this annoying you, please just assume it is defined piecewise such as the function is zero in the undefined point.

This comment is important, since I want to focus the discussion in the other singularities that appears on these equations (and sometimes the point could be coincident), and are the discontinuities on the differential equations due the terms $(T-t)$.

The questions

1) Are all these differential equations with terms $(T-t)$ "well-defined" as differential equations?

2) Which are the name of this kind of differential equations? (noting here there are ODEs and PDEs with these terms - hope there are any references)

3) Does this $(T-t)$ terms on the denominator make these differential equation Non-Lipschitz? Here looking for parallels with the papers by V. T. Haimo: Finite Time Differential Equations and Finite Time Controllers.

4) Does this $(T-t)$ terms on the denominator make possible the existence of solutions of finite duration? Remembering here that solutions of finite duration don't fulfill uniqueness (see the recently cited papers for better explanation), a singularity in my opinion makes sense since there are multiple values rising from a single point, which is similar with traditional mistakes made when dividing by zero - but formally speaking, I have no idea of what it is happening.


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 (I still didn't find a related framework).

Current physics solutions 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, which at least under the assumption that thermal noise is Gaussian, you can say if it still moving (neither if it has stopped), 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 (see the comments of this question to a more detailed discussion applied to the pendulum movement).

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).

But nowadays there is a lot of theoretical physics squeezing their models to the very last drop and many counter-intuitive things are reaching mainstream: many Youtube videos talking about the "nonexistence" of time (is an illusion), and also many videos about the impossibility of current physics laws to determine "the direction of the arrow of time", and I would like to know if any of this things is due current models are not aware of the effects of keeping the assumption that phenomena have finite ending times.

As example, in all the differential equations of this question there indeed a clue of which way the time is considered to be flowing (due the term $(T-t)$), and for the solutions, going backwards in time will lead to spontaneous rise from zero (being zero from a non-zero meassure interval in time).

I know that current models are on the way they are because the models were designed to keep holding uniqueness of solutions, which if not could lead to a lot of struggles (see here some of this issues with the Norton's Dome example).

But I also know now that for the existence of solutions of finite duration uniqueness of solutions must be abandoned, and I would like to know if this were known by physicists, if the current models would still been having their current formulations (to keep this on perspective, main models have about a hundred years old, and the first papers about finite duration solutions I have found are from 1985).

And maybe because they are "relativately new", I only found these finite duration solutions on control theory papers (where they started assuming a high-knowledge background from the reader), and every book I find on differential equation keep their analysis tied to uniqueness, so they don't touch this topic. With this, any related book reference will be useful.

Answering two firsts comments by @Desura

Some user named @Desura mentioned the existence of the Frobenius method and the Fuchs' theorem for solving differential equations of the form $u''+\frac{p(t)}{t}u'+\frac{q(t)}{t^2}u=0$, and also what is a Regular singular point. On all these pages you can see differential equations that really resemble the ones I have displayed above, so very much for share it.

I would like to comment why I think these kind of methods aren't able to find solutions of finite duration by "themselves", and also how I think I am already using them through Wolfram-Alpha.

On the mentioned papers by Vardia T. Haimo (1985) Finite Time Controllers and Finite Time Differential Equations, where only scalar autonomous ODEs of 1st and 2nd order are studied ($x'=G(x)$ or $x''=G(x',x)$), then by assuming without lost of generality that $T=0$, and also assuming that the right-hand-side of the ODE is at least class $C^1(\mathbb{R}\setminus\{0\})$, the author mentioned that:

"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, if the author is right, NO Lipschitz 1st or 2nd order scalar autonomous ODEs could stand solutions of finite duration, so classic linear ODEs are discarded.

But reaching zero in finite time have also another consequence, and is that also are discarded classical solutions through Power Series which are analytical in the whole domain, since there exist a non-zero measure compact set of points identical to zero, the only analytical function that could support this matching is the zero function, due the Identity Theorem.

Because of this last paragraph, is why I believe that a method that find the coefficients for a function of the form $\sum_k a_k(T-t)^k$ aren't able to describe a finite duration solution "alone", which is a traditional method for finding solution to ODEs (as the method shown by @Desura).

But fortunately, being analytic in a piecewise part of the time domain and zero on the part "looks like" working if the two restrictions are met (now labeled after \eqref{Eq. 7}), as they are used for finding the solution of the now labeled \eqref{Eq. 14}, which is demonstrated on this answer. Here is where I am using the traditional ways of finding solutions to differential equations, to find the "non-zero section".

But How this "construction" by "stitching solutions" is formally working, or if they are being or not formal solutions to the differential equations is something I don´t know (I am "highly abusing" of an answer without any formal prove - but because it "make sense"), neither when we have to "choose" one solution or another (since there are multiple options now - at least with this "construction": the full solutions, the finite duration alternative, or the other piece that rise spontaneously from zero). This is why I am looking for a framework, but because as requirement uniqueness must be left aside, traditional books in differential equations don't touch this kind of solutions.

Even so, on the mentioned papers of V. T. Haimo, this kind of solutions aren't studied at all, and more interestingly, there are given a 2nd order ODE example where is shown that depending on its parameters, the differential equation could allow having finite duration solutions, but outside "this restriction" the solutions are not of finite duration (unfortunately no exact solution is given).

Last update

I have just found the following papers under the term sublinear damping that shows example in physics of equations really similar to one showed on the papers by V. T. Haimo, containing terms of the form $\text{sgn}(\dot{x})|\dot{x}|^\alpha$ which are achieving finite extinction times without having the form of the Frobenius method with displaced terms $(T-t)$:

I don't know if maybe after knowing their exact solutions, they could be transformed on differential equations similar of the asked on the main question, as it was done for carrying \eqref{Eq. 2} to become into \eqref{Eq. 3} (I believe implying with this their decaying is exponential $\propto t^{-\beta},\,\beta>0$ for some $\beta$), maybe it could be proved with what is shown on the papers. Other similar examples were listed into this another question.

  • 1
    $\begingroup$ You might be interested in studying first the following class of linear $n$-th order ODEs $f^{(n)}(z) + \sum_{i=0}^{n-1} p_i(z)f^{(i)}(z) = 0$ with initial data at $z_0 \in \mathbb C$ and coefficients $p_i(z)$ which are meromorphic around $z_0$ for all $i \in \{0,1,...,n-1\}$. Then the point $z_0$ can be classified in three ways : an ordinary point, a regular singular point and a fully irregular point. $\endgroup$
    – Desura
    Commented May 24, 2022 at 20:05
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    $\begingroup$ The existence and uniqueness theory breaks only in the fully irregular case. In the regular singular case, solutions with singularity at $z_0$ can exist but this is not always the case and it will depend only on the poles of the $p_i(z)$'s (look for Frobenius Theorem, Fuchs Theorem or Fuchsian Differential Equations). Please note that your equation (7) has a regular singular point at $z_0 = 2$. $\endgroup$
    – Desura
    Commented May 24, 2022 at 20:05
  • $\begingroup$ @Desura thanks you very much for your interesting comments. I didn't know before about the Frobenius method neither about Regular singular point. I will add, bacause of extension, what I think into the question during the next hour. $\endgroup$
    – Joako
    Commented May 24, 2022 at 21:15
  • $\begingroup$ @Desura I have just upload on the question my comments. Hope you can elaborate after read them. Beforehand, thanks you very much. $\endgroup$
    – Joako
    Commented May 25, 2022 at 1:43
  • $\begingroup$ It seems to me that the sentence phenomena have finite ending-times is ambiguous. For example, the autonomous ODE $y^\prime = y^2$ has solutions exploding to infinity in a finite time. So formally, the system has a finite ending time. Is it something that you regard as a phenomenon that has a finite ending-time? IMHO, I think that you should revisit your question by trying to precisely define what is still fuzzy. That will help yourself to understand in more detail the topic... And then facilitate answers! $\endgroup$ Commented May 25, 2022 at 10:22

2 Answers 2


1) Are all these differential equations with terms $(T-t)$ "well-defined" as differential equations?

Their well-posenedness will depend on the multiplicity of the singularities. You will need to check whether the singular points are regular or irregular. See e.g. https://en.wikipedia.org/wiki/Regular_singular_point. You may also check the Frobenius method which will be particularly useful for some of your differential equations: https://en.wikipedia.org/wiki/Frobenius_method.

2) Which are the name of this kind of differential equations?

They are commonly referred to as singular differential equations.

3) Does this $(T-t)$ terms on the denominator make these differential equation Non-Lipschitz?

Of course, they make it non-Lipschitz. A necessary condition for being Lipschitz is to be continuous and $1/(T-t)$ is not continuous at $t=T$.

4) Does this $(T-t)$ terms on the denominator make possible the existence of solutions of finite duration?

It may but that may not be the best solution to do so. You gave the example in Eq. 4 and Eq. 5.

  • 1
    $\begingroup$ Of course, they make it non-Lipschitz. A necessary condition for being Lipschitz is to be continuous and $1/(T−t)$ is not continuous at $t=T$. The problem is not so much that $1/(T−t)$ is not continuous at $t=T$. It is that it is not even defined! However $t \mapsto 1/(T−t)$ is locally Lipschitz for any $t$ where the map is defined. Which is enough to ensure that the ODE has a unique maximal solution at any point $t_0 \neq T$. $\endgroup$ Commented May 25, 2022 at 14:05
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    $\begingroup$ @Joako That's an interesting problem. $\endgroup$
    – KBS
    Commented May 25, 2022 at 19:17
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    $\begingroup$ @Joako there has been a lot of work on the topic in control theory for finite convergence of control systems and the like. This has been well-studied. $\endgroup$
    – KBS
    Commented May 25, 2022 at 19:48
  • 1
    $\begingroup$ You will almost never find explicit solutions to differential equations. Most of them are not explicitly solvable only the simple ones are. What is important are the qualitative properties (e.g. stability) and certain quantitative properties (e.g. a bound on the finite convergence time). $\endgroup$
    – KBS
    Commented May 25, 2022 at 20:14
  • 1
    $\begingroup$ What is $g(x)$? $\endgroup$
    – KBS
    Commented May 26, 2022 at 0:05


In what follows, we'll only consider first-order Ordinary differential equations (ODE) defined as $$y^\prime(t) = f(t, y(t))$$ where $f : U \to \mathbb R$ is continuous and $U\subseteq \mathbb R^2$ is an open subset.

According to Peano existence theorem, the Initial value Problem (IVP)

$$\begin{cases} y^\prime(t) &= f(t, y(t))\\ y(t_0) &= y_0\end{cases}$$ where $(t_0,x_0) \in U$ has at least one maximal solution.

Interestingly, the solution may not be global as can be seen with the case $$\begin{cases} y^\prime(t) &= y^2(t)\\ y(1) &= 1\end{cases}$$ that explodes to infinity in a finite time. By the way an interesting theorem states that either the solution of such ODEs is global, either it reaches points outside of any compact in a finite time.

The solution may also not be unique as can be seen with the following example$$\begin{cases} y^\prime(t) &= \sqrt{\lvert y(t) \rvert }\\ y(0) &= 0\end{cases}$$

However, as stated by Picard–Lindelöf theorem, the IVP $$\begin{cases} y^\prime(t) &= f(t, y(t))\\ y(t_0) &= y_0\end{cases}$$ has a unique maximal solution for any $(t_0,y_0) \in U$ providing that $f(t, y)$ is locally Lipschitz in $y$ for any $y_0$. This is in particular the case if $f$ is continuously differentiable.

Solutions of finite duration

So coming back to your question, you're expecting to get solutions $y$ defined in an open interval $(a,b)$ for which $y(t) = 0$ for $t \ge T$ where $T \in (a,b)$ and... that are not always vanishing. That would be too simple.

You can't get such solutions according to Picard–Lindelöf theorem if $f$ is locally Lipschitz in $y$ around $y(T)$.

But you don't need to have $t-T$ in the denominator of some term of the ODE to get the finite duration phenomena. $$\begin{cases} y^\prime(t) &= \sqrt{\lvert y(t) \rvert }\\ y(0) &= 0\end{cases}$$ is a good example for that!

I know that this doesn't answer to your precise 4 questions. However, I think that this provides you with a much global way to study the finite duration topic you raised.

Interesting would be to answer following question.

Suppose that you have an autonomous IVP $$\begin{cases} y^\prime(t) &= f(y(t))\\ y(t_0) &= y_0\end{cases}$$ where $f : \mathbb R^2 \to \mathbb R$ is continuous, continuously differentiable for $y \neq 0$ with $f(0)=0$ and $\lim\limits_{y \to 0} \lvert f^\prime(y) \rvert = \infty$. Does we then have finite duration solutions? This would give a very general case of finite duration solutions.

  • $\begingroup$ Thanks you very much for taking the time for this this interesting insight. I don't fully get the last part: you are asking for a finite duration solution with an discontinuity in the acceleration profile at $t=0$?, If this is the case made a similar question here and find that something of the form $$x(t) = \frac{(1-t^2+|1-t^2|)\,t\log(t^2)}{4}\,\exp\left(-\frac{t^2}{1-t^2}\right)$$ could do that trick. $\endgroup$
    – Joako
    Commented May 26, 2022 at 18:26
  • $\begingroup$ What I am doing in the last part is to try to generalize the case $y^\prime =\sqrt{\lvert y \rvert}$ to get a more general condition on which finite duration solutions exist. For this, I tried to identify what is specific to this particular example. $\endgroup$ Commented May 26, 2022 at 18:44
  • $\begingroup$ If is the generalization of a scalar autonomous IVP, is the first case studied on the papers by V. T. Haimo a cited on the question above... I am not sure if matched your definition at 100% (because of the $\mathbb{R}^2$), Did you review them? I recommend you to see first "Finite Time Controllers", since the other have an important typo mistake in an inequality sign. $\endgroup$
    – Joako
    Commented May 26, 2022 at 18:52
  • $\begingroup$ On V. T. Haimo papers are shown required conditions, but I believe now that the two conditions I explain above are sufficient: if (i) zero is solution then $y''=F(y',y)$ leads to $F(0,0)=0$; and (ii) having a singular point $-\infty<T<\infty$ where $y(T)=y'(T)=0$ leads to $y''(T)=F(y'(T),y(T))=F(0,0)=0$, and since is singular there uniqueness could be broken, so conditions to stich solutions with the zero function continuously are met. Same can be done for the case $y'(t)=F(y(t))$. $\endgroup$
    – Joako
    Commented May 27, 2022 at 17:51
  • $\begingroup$ Other interesting thing from the paper of V. T. Haimo, is that in her analysis she found that for the differential equation: $$\ddot{x}=-\text{sgn}(x)|x|^a-\text{sgn}(\dot{x})|\dot{x}|^b$$ founding that if $0<b<1$ and $\frac{b}{2-b}<a$ then there are solutions of finite duration. Here, completely different from the two conditions analysis I did above. $\endgroup$
    – Joako
    Commented Jun 2, 2022 at 23:15

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