Take any differential equation of the form


where $n > 1$. The solution $y(x)$ will reach infinity at a finite value of $x$.

Assuming $y_0 =1 $ for all cases, here are a few examples: $$\frac{dy}{dx}=y^2$$ has the solution $$y=\frac{-1}{x-1}$$
which reaches its asymptote at $x=1$.

The DE $$\frac{dy}{dx}=y^{1.01}$$ has the solution $$y=\left(\frac{-100}{x-100}\right)^{100}$$
which reaches its asymptote at $x=100$.

If you take any DE of the form $$\frac{dy}{dx}=y^{1 + \epsilon}$$ where $\epsilon$ is a very small number, the solution is $$y=\left(\frac{-1}{\epsilon(x-\frac{1}{\epsilon})}\right)^{\epsilon^{-1}}$$
which eventually hits the vertical asymptote at the very large number $\frac{1}{\epsilon}$

This has always bugged me. Intuitively, one expects that the solutions to these equations will grow rapidly and aggressively, much faster than the exponential function. But it is not entirely obvious why they should reach an infinite value after a finite time, instead of say, grow like the Ackermann function or some other function that grows rapidly but stays strictly finite.

Is there an intuitive argument for why these DEs are able to reach infinity in a finite timespan?

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    $\begingroup$ Sure... for the same reason $f(t) = 1/(t-1)$ goes to infinity in a finite time. $\endgroup$ Dec 5, 2019 at 0:58
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    $\begingroup$ I'd like to mention (though it's perhaps irrelevant) that this can happen in Newtonian gravitation: en.wikipedia.org/wiki/Painlev%C3%A9_conjecture $\endgroup$
    – mr_e_man
    Dec 5, 2019 at 2:33
  • $\begingroup$ Runaway feedback. Think of a screetching microphone getting louder and louder. More of a handwave than an intuition perhaps, but things don't have to go to infinity to get out of control. $\endgroup$ Dec 6, 2019 at 0:04
  • $\begingroup$ E.g. the $1/(1-x)$ example. The singularitiy might be a finite distance away. But are you really able to reach it? Probably not in some physical situation. In physics infinite quantities are usually not considered realistic. And mathematically I would consider $y(x=1)$ undefined. $\endgroup$
    – mvw
    Dec 6, 2019 at 15:40
  • $\begingroup$ Thinking about the Euler's Disk toy which wobbling rate rise as a finite time blow up, maybe is due the denominator is behaving as a finite duration solution, which becomes zero by its own dynamics and stays there forever. I just made a related question trying to figure out the relation among finite duration solutions as the reciprocal of blow ups solution. I think it could be an intuitive approach for some models (as is noted in the first comment, is not always true). $\endgroup$
    – Joako
    Apr 7, 2022 at 23:18

5 Answers 5


The point is that $dy/dx = y^p$ is equivalent to $dx/dy = y^{-p}$, i.e. instead of thinking of $y$ as the dependent variable and $x$ as independent, do the reverse. If you think of $x$ as position and $y$ as time, the velocity is $y^{-p}$. If $p > 1$, this goes to $0$ fast enough that the change in $x$ as $y$ goes from some finite positive value to $\infty$ is finite. Now change point of view again and it says that as $x$ goes to some finite value, $y$ goes to $\infty$.


Your intuition that a solution to a DE like this should grow quickly but finitely makes a lot of sense. One justification for this intuition is to look at the estimation Euler's method would give: entirely finite and defined for the whole real line. To fix this inaccurate intuition, consider the following improvement of Euler’s method: instead of increasing x a constant amount each time, only increase x far enough to let y double. Since y doubles with each jump, $y^n$ Increases by $2^n$, so the ratio of the size of the horizontal jump from one jump to the next decreases by a factor of $\frac {2}{2^n}$. since n>1, this ratio is less than one. As a result the x-position converges, so y is doubling with out bound but x converges.

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    $\begingroup$ Xeno's D.E. :-) $\endgroup$ Dec 5, 2019 at 19:47
  • $\begingroup$ @DukeBouvier what does that mean? $\endgroup$ Dec 11, 2019 at 13:57
  • $\begingroup$ Xeno's paradox: to move ten yards, I must first go 5 yards to half way. And another 2.5 yards, another 1.25 years, ad infiniturm. Without the idea of an infinite sequence adding to a finite sum, Xeno's paradox was that you could never move because it takes an infinite number of steps - so (it was said) would take an infinite time. If I understand correctly, @Robo300 is suggesting that the same flawed initution is at work in the suggestion that an infinite sequence of steps along the x-axis must take us to x=infinity rather than to the asymtotic finite maxmium. $\endgroup$ Dec 12, 2019 at 17:12
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    $\begingroup$ @Duke Bouvier, that wasn’t exactly the flawed intuition I had in mind. If you go about the more traditional way with Euler’s method, you come up with a process that really does take infinite “time” , behaving much like the question suggests and growing at an enormous bit finite rate. My point was to take a more careful estimation, where it becomes clear that the process reaches infinity in finite time. But I do like your explanation too $\endgroup$
    – Robo300
    Dec 12, 2019 at 18:12

There is a nice discussion of this problem here p. 423, where the authors show by example that what one expects is not necessarily what happens. Below is a sketch of their proof of a criterion which can be used to tell whether a solution will blow up in finite time. Namely, we have a


if $y'=f(y);\ y(0)=y_0;\ f(y)>0$ for all $y>y_0,$ then $y$ blows up at time $t_1$ if and only if $\int^{\infty}_{y_0}\frac{1}{f(y)}dy=t_1.$

For the proof, note that $\int^{y(t)}_{y_0}\frac{1}{f(u)}du=t$ whenever the integral is defined. Therefore, if $y$ satisfies $\underset{t\to t_1^-}\lim y(t)=\infty$ then $\underset{t\to t_1^-}\lim \int^{y(t)}_{y_0}\frac{1}{f(u)}du=\underset{t\to t_1^-}\lim t=t_1.$

On the other hand, if the integral converges to $t_1,$ then $t=\int^{y(t)}_{y_0}\frac{1}{f(u)}du<\int^{\infty}_{y_0}\frac{1}{f(u)}du=t_1$ so $t$ is bounded by $t_1$. Conclude by observing that

$\underset{t\to t_1^-}\lim \int^{y(t)}_{y_0}\frac{1}{f(u)}du=\underset{t\to t_1^-}\lim t=t_1=\int^{\infty}_{y_0}\frac{1}{f(u)}du$ so $\underset{t\to t_1^-}\lim y(t)=\infty.$

  • $\begingroup$ Did you see the comment on the other answer? $\endgroup$
    – mr_e_man
    Dec 5, 2019 at 2:38
  • $\begingroup$ I just read it. It appears the OP is not getting the answer he/she wants. $\endgroup$ Dec 5, 2019 at 2:54

Its because $\int_1^\infty \frac{1}{y^p} dy$ is finite for $p>1$, but infinite for $p=1$.

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    $\begingroup$ This isn't really the sort of answer I'm looking for. There are many possible variations on "it's infinite because that's what the equation says". I'm looking for an answer that takes the DE at face value and explains how something can grow so fast that it can become infinite in finite time. $\endgroup$
    – Ingolifs
    Dec 5, 2019 at 2:30
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    $\begingroup$ That is exactly what I did. $\int dx = \int dy/y^p$; if the right hand side is finite then so is the left hand side. But I understand if this isn't what you are looking for. Its difficult to get a personally satisfying answer to such questions. $\endgroup$
    – Spencer
    Dec 5, 2019 at 2:40
  • $\begingroup$ But $any$ function with a vertical asymptote at $t=t_0$ "becomes infinite" in finite time so maybe I do not understand what you are asking. $\endgroup$ Dec 5, 2019 at 2:52
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    $\begingroup$ So, Imagine you're numerically integrating the DE on a computer with arbitrarily large floating point numbers. Intuitively, you'd expect the solution would grow very rapidly, but still always stay finite. Now reduce the step size. Maybe you'd find the solution grow much faster then before but still stay finite. The proof I had in my mind that I'm not clever enough to put on paper would be able to describe the expected behaviour once the step size becomes infinitely small. I'm sure that's mathemetically equivalent to what you said but I don't see it yet. $\endgroup$
    – Ingolifs
    Dec 5, 2019 at 2:55

I thought I'd come back to this and give an answer that satisfies the original intent of the question as I asked it.

Differential equations are infinitely responsive.

The differential equation reacts infinitely fast to any changes in the equation. Simply put, this allows for the possibility of an infinite amount of change occurring in a finite period. If you alter $a$ in $a \frac{dy}{dt}$, you'll see a change instantaneously, at time $t$. This is a major point of difference to numerically simulated DEs, because the action of $dy/dt$ only takes effect at $t +\Delta t$.

Physical reality is not infinitely responsive.

Every force or interaction between one particle and another in this universe can only propagate - at maximum - at the speed of light. In the vast majority of cases, the interaction propagates much more slowly than this.

I guess the issue for me came when conflating 'differential equations as a tool for understanding the physical universe', with 'differential equations as a mathematical construct that are under no obligation to make physical sense'.

If you numerically approximate $dy/dx = y^2$ using the Euler method on your laptop, you will indeed see $y$ rapidly increase beyond the confines of the 32-bit floating point format, but $y$ will still always stay finite. It's easy to prove this - The difference between timesteps is finite, therefore its square in the next timestep will be finite.

When you look at the role differential equations play in physics, they are often in the context of treating an uncountably large number of particles as a continuous object. Once the scales become smaller than individual particles, the assumptions will become unphysical, and so will the results of the DE.


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