Any physical example of an "explosive" differential equation $ y' = ky^2$?

Question

I was told that in physics (and in chemistry as well) there are processes that may be described by a differential equation of the form $$ y' = ky^2. $$ That is, the variation of a variable depends from the number of pairs of the elements.

I understand the mathematical meaning of that equation, but I cannot conceive of any physical process which leads to this. Maybe some nuclear reaction could, but I am at a loss. Any hints?

A "drag" force that is used to approximate motion of a body in a fluid, (i.e $F=kv^2$) is the first example that comes to my mind. — Antillar Maximus, Jul 26 '13 at 22:17

score 15 · Answer 1 · edited Jun 11 '20 at 09:33

Your equation is a special case of Riccati equation: $$y'=q_0(x) + q_1(x)y + q_2(x)y^2\!$$ with $q_0(x)=q_1(x)=0$ and a constant $q_2(x)$: $$y'= ky^2\!$$ There are lots of applications for the main Riccati equation in physics, and some of them can be reduced to the special case of $y'= ky^2\!$. (Although, explosive behavior is usually avoided and means failure of the model)

There are some beautiful examples in the context of cosmology, early universe and expansion of the universe that give rise to Riccati equation.

One example is Friedman equation(s), with an equation of state of the form $p=\omega \rho$ with a constant $\omega$: (c is a constant, not the speed of light, speed of light=1 in this system of units) $$\cases{\ddot{R}=-\frac{4\pi G(\rho+3p)R}{3} \\ \dot R=\frac{8\pi G\rho R^2}{3}-k}$$ with the following assumptions: $$c=\frac{1-3\omega}{2}\,\,\,\,\,\,R(\eta)d\eta=dt\,\,\,\,\,\,u=\frac{1}{R}\frac{dR}{d\eta}$$ we will arrive at the following Riccati equation: $$\frac{du}{d\eta}+c(\eta)u^2+kc(\eta)=0$$ For a flat universe k=0; and with the substitution $u=\frac{1}{c}\frac{y'}{y}$ we can transform it to a second order ODE, which has a solution of the form $R(\eta)=(a\eta) ^{\frac{2a}{1-3\omega}}$ which is an expanding universe. ($a$ is a constant) For other cosmological examples see this.

There is a paper that has some other interesting examples; I bring some of them here: http://arxiv.org/abs/physics/0110066

Quadratic friction

Consider the quadratic friction law ($f=-\alpha v^2$) for an object with mass m. With the external force being zero the Newton's second law for the object will be: $$\dot v=-\alpha v^2$$

Central Potential of the form $V(r)={k\over r^2}$

Assuming a zero total energy for a particle (so $k<0$), equation of motion for the particle under the potential $V(r)=kr^{\epsilon}$ will be of Riccati form. Angular momentum and energy are conserved quantities:

$$T=\frac{1}{2}mv^2=\frac{1}{2}m{\dot r}^2+\frac{1}{2}mr^2\omega^2=\frac{1}{2}m{\dot r}^2+\frac{l^2}{2mr^2}$$ $$E=T+V(r)$$ $$\cases{E=0\\\dot E=0}\to$$ $$\frac{1}{2}m{\dot r}^2+\left( \frac{1}{2}+\frac{1}{\epsilon}\right)\frac{l}{mr^2} - \frac{m\ddot r r}{\epsilon}=0 $$

Now if you write the above equation in polar form (i.e., $r=r(\theta(t))$ ) you'll arrive at a Riccati equation: $$\dot{\theta}=\frac{l}{mr^2}\,\,\,\,r'={dr\over d\theta} \,\,\,\,\omega={r'\over r}\to$$

For the general case $$\omega '=\frac{\epsilon +2}{2}\omega^2 +\frac{\epsilon +2}{2}$$ For $\epsilon=-2$ the equation will be of $y'= ky^2\!$ form!

score 13 · Answer 2 · answered Jul 26 '13 at 11:54

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If I have $n$ objects (say, reactants) colliding, the rate of collisions will be roughly proportional to $n^2$. If the population grows by a fixed amount with each collision, we would find this law. See the rate equation. I think that growth due to sexual reproduction might fit here as well, but I'm not familiar enough with population biology to say.

answered Jul 26 '13 at 11:54

Robert Mastragostino

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I see this in something called Target Mediated Drug Disposition, a two-input one-output reaction, although it's not explosive because it runs both ways.

Mike Dunlavey

Jul 26 '13 at 12:33

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I don't think biological population growth would fit. Except at very low levels (where just finding a mate is problematic), doubling the total population won't cause each female to have twice as many offspring. – Dan Is Fiddling By Firelight Jul 26 '13 at 13:53

@DanNeely what I was thinking is along these lines: I have a population of $n$ animals, equally male and female, randomly roaming around. If a male and a female run into each other (which should happen with frequency $n^2/2$), the population increases by some fixed number (say, 2 offspring). Are you saying this isn't an example, or that it's just unrealistic? – Robert Mastragostino Jul 26 '13 at 17:05

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@RobertMastragostino The population growth rate will asymptote to $\mathcal{O}(n)$ so long as the gestation period extends over a greater-than-infinitesimal time. $\mathcal{O}(n^2)$ growth only happens in the limit that mating is limited by the frequency of encounters with males. – Jul 26 '13 at 17:21

@RobertMastragostino What Chris White said. Except when population densities are very low it's not relevant. – Dan Is Fiddling By Firelight Jul 26 '13 at 17:30

score 11 · Accepted Answer · edited Jul 26 '13 at 16:50

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Well, simpler than recognising a differential equation will be to recognise the quantity it describes. Quickly plugging in Solve[y'[t]==k y[t]^2,y[t],t] into wolfram alpha gives

$$y(t)=\frac{1}{c-k\ t},$$

which explodes at $t=c/k$. Now let's think of physical quantities which diverge as $\frac{1}{t}$. The Coulomb potential

$$\phi(r) = \frac{Q}{4 \pi \varepsilon_0} \frac{1}{r_0-r} = \frac{1}{\left(4 \pi \varepsilon_0Q^{-1}\ r_0\right)-\left(4 \pi \varepsilon_0Q^{-1}\right)\ \cdot\ r}$$

of a point charge $Q>0$ at $r_0$ comes to mind. We position ourselves at the one side, $r<r_0$, where the function behaves as $\frac{1}{-r}$ and so is positive. Its derivative is $$\left(\tfrac{1}{-r}\right)'=\tfrac{1}{r^2}=\left(\tfrac{1}{-r}\right)^2,$$ which is the square. So if you approach a charge, its potential value changes as $\phi'(r)\propto \phi(r)^2$.

edited Jul 26 '13 at 16:50

AJMansfield

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answered Jul 26 '13 at 12:24

Nikolaj-K

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But do you think there is any deeper physical significance to the fact that the Coulomb potential happens to fulfill that differential equation? Does that even make sense in three-dimensional space? It can't be very insightful to discuss this potential in only one dimension, and in $\mathbb{R}^3$ (and only there!) it follows very naturally from Gauss' law, as an integral rather than differential equation. – leftaroundabout Jul 26 '13 at 22:33
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@leftaroundabout: It holds if you approach the charge in a straigth line. The 1 dim is simply the radius. And the question of "deep physical significance" is always vauge - things are describably by some math and this entails many things. You can take the time derivative of $F(r(t))=mr''(t)$ and get some new differential equation whos significance I wouldn't want to defend. The point is that it has not much practical use. Similarly, the differential equations of electrostatics let you find $\phi$ via the charge $\rho$. Truths about the solution, like $\phi'\propto\phi^2$, are secondary. – Nikolaj-K Jul 27 '13 at 00:17
@leftaroundabout An expression as simple as $y(x) = 1/x$ (like the Coulomb potential) will satisfy many simple differential equations. The vast majority of them will not have any "deeper physical significance". – tparker Oct 01 '17 at 02:21

score 1 · Answer 4 · answered Oct 01 '17 at 02:29

This is perhaps stretching the definition of "physical" but Metcalfe's law says that the value of a telecommunications network is proportional to the square of the number of users. A growing network might reasonably have a rate of growth proportional to its "value", which would lead the the differential equation you describe, so some networks might grow hyperbolically. See https://en.wikipedia.org/wiki/Hyperbolic_growth#Applications for more examples.

score 1 · Answer 5 · answered Jul 26 '13 at 16:10

The population growth of an imaginary species with two sexes, where childbirth and growing up is instantaneous and there's no death.

A dictator decrees that every man and woman must mate, on average, once a year, with each mating producing a child (of random sex). So if there are 1000 men alive today, each woman has to have 1000 children in the next year, one with each man. But it's actually much more than 1000, because of all the male children born during the course of the year...

score 1 · Answer 6 · answered Jul 26 '13 at 17:09

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As described in Wikipedia, world population was growing in just that fashion (hyperbolic) for some time in the 20th century.

answered Jul 26 '13 at 17:09

user27542

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Any physical example of an "explosive" differential equation $ y' = ky^2$?

6 Answers6

Quadratic friction

Central Potential of the form $V(r)={k\over r^2}$