18
$\begingroup$

Take any differential equation of the form

$$\frac{dy}{dx}=y^n$$

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 $x=100$.

and 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?

$\endgroup$
  • 4
    $\begingroup$ Sure... for the same reason $f(t) = 1/(t-1)$ goes to infinity in a finite time. $\endgroup$ – David G. Stork Dec 5 '19 at 0:58
  • 10
    $\begingroup$ 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 '19 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. $\endgroup$ – John Coleman Dec 6 '19 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 '19 at 15:40
20
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ Xeno's D.E. :-) $\endgroup$ – Duke Bouvier Dec 5 '19 at 19:47
  • $\begingroup$ @DukeBouvier what does that mean? $\endgroup$ – Lukas Juhrich Dec 11 '19 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. $\endgroup$ – Duke Bouvier Dec 12 '19 at 17:12
  • 1
    $\begingroup$ @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 '19 at 18:12
23
$\begingroup$

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 $\infty$.

$\endgroup$
3
$\begingroup$

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

Theorem:

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 $\underset{t\to t_1^-}\lim y(t)=\infty.$

$\endgroup$
  • $\begingroup$ Did you see the comment on the other answer? $\endgroup$ – mr_e_man Dec 5 '19 at 2:38
  • $\begingroup$ I just read it. It appears the OP is not getting the answer he/she wants. $\endgroup$ – Matematleta Dec 5 '19 at 2:54
2
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ 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 '19 at 2:30
  • 3
    $\begingroup$ 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 '19 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. $\endgroup$ – Matematleta Dec 5 '19 at 2:52
  • 1
    $\begingroup$ 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 '19 at 2:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.