Blow-up in finite time

Example

Let $x_0 > 0$ and consider the initial value problem $x'(t) = x(t)^2$, $x(0) = x_0$. Separation of variables gives the solution $$x(t) = \frac{x_0}{1 - x_0 t}, \qquad t \in \left(-\infty, \frac{1}{x_0}\right).$$ This is the maximal solution: $x(t) \to \infty$ as $t \to 1/x_0$ from below. So the solution blows up at the finite time $1/x_0$, although the field $f(x) = x^2$ is smooth on all of $\mathbb{R}$.

Remarks

The blow-up time $1/x_0$ depends on the initial value. Larger initial values blow up sooner. For $x_0 < 0$, the maximal interval is $(1/x_0, \infty)$, and for $x_0 = 0$ the solution is $0$ on $\mathbb{R}$. The field $x^2$ is not globally Lipschitz. A globally Lipschitz field has no blow-up. Near the blow-up time, $x(t)$ behaves like $(1/x_0 - t)^{-1}$. This is a finite-time singularity.

Explanations

L1 human

The rate of growth is the square of the size, so the growth feeds itself very fast. The solution reaches infinity in finite time, unlike exponential growth.