# Estimating rate of blow up of an ODE

Suppose I have a differential equation $x'=f(x)$ and $f(x)>0$ grows super-linearly. I.e., $\lim_{|x| \rightarrow \infty} |f(x)|/|x| \rightarrow \infty$.

Several related questions: (1) Can I conclude blow up of a solution for some initial conditions? (2) Via taylor series, or whatever relevant approximations of $f(x)$, can we estimate the rate of blow up? And (3), again by approximation, can we estimate given an initial condition when this blow up occurs?

I have in mind $x'=x^p$ which has solution $x=[(1-p)t-C]^{-1/(p-1)}$. There is finite time blow up of order $O(t^{1/(p-1)})$. Given an initial condition $x(0)=X$, we have blow up occur at $X^{1-p}/(p-1)$.

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With initial condition $x(0)=X$, the solution to your differential equation satisfies $$\int_X^{x(t)}\frac{dy}{f(y)} = t,$$ as one can check by differentiating both sides and using the chain rule. The necessary and sufficient condition for blow-up at a finite time $T$, meaning $x(t)\to\infty$ as $t\to T$ from below, is then that $$\int_X^\infty \frac{dy}{f(y)}=T.$$

An explicit asymptotic estimate is available for $T$ as $X$ becomes large if $f$ is "regularly varying with exponent $p>1$". This means that $f$ is almost a power law, in the sense that $$f(y)=y^p L(y) \quad\mbox{ where }\quad \frac{L(cy)}{L(y)}\to 1$$ as $y\to\infty$ for every $c>0$. For then one can deduce that as $X\to\infty$, the blow-up time $$T = \int_X^\infty \frac{dy}{y^p L(y)} = \frac{X}{X^pL(X)}\int_1^\infty \frac{L(X)}{L(Xz)}\frac{dz}{z^p} \sim \frac{X^{1-p}}{L(X)} \frac{1}{p-1}.$$ To justify the last step one needs to use some basic facts proved in books on regularly varying functions, such as those of Seneta or Bingham, Goldie and Teugels. In particular, (i) the convergence $L(X)/L(Xz)\to1$ occurs uniformly for $z$ in any compact set, and (ii) there is a uniform bound $L(X)/L(Xz)\le Cz^\epsilon$ independent of $X$, for any $\epsilon>0$.

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The conditions you have for 'superlinearity' are not sufficient for finite-time blowup. For example, $x' = x \log x$ with initial condition $x(0) > 1$. This equation grows superlinearly, and yet does not exhibit finite-time blowup (to solve it, change variables to $y = \log x$ and solve $y' = y$, then back-substitute).

The idea is vaild, however, if you allow a little more growth (for e.g. $|f(x)|/|x|^{1+\epsilon} \rightarrow \infty$ for some fixed $\epsilon > 0$). Then you could conclude finite blowup time by comparing the solution to $x' = f(x)$ with the solution to $x' = |x|^{1+\epsilon/2}$, for example, which does blow up.

I would be careful with the Taylor series. Especially if $f > 0$, your solution is likely to increase indefinitely, in which case terms arbitrarily far out in the taylor expansion are going to dominate the dynamics. You might be safe, though, if all the coefficients in the expansion are positive.

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Nice counterexample! – nayrb Feb 1 '13 at 22:08