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

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

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!

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.

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 $\phi'(r)\propto \phi(r)^2$.