I am faced with a non-autonomous initial value problem for a function $x:[0,\infty) \to \mathbb{R}^2$ of the form $$ x'(t) = f(t,x(t)) $$ for $f: [0,\infty) \times \mathbb{R}^2 \to \mathbb{R}^2$ with initial condition $x(0)$. Now, the function $f$ is such that $$ f(t,x) = t^{-1} f_{-1}(x) + f_0(x) + t f_1(x) $$ where the functions $f_n: \mathbb{R}^2 \to \mathbb{R}^2$ are analytic (in fact, polynomial). Furthermore, the initial condition is such that $$\lim_{t\to 0^+} f(t,x(t))$$ exists. This means that there is a formal power series for $x(t)$ around $t=0$ which solves the initial value problem; although it depends on a parameter (related to $x'(0)$) which cannot be fixed. In other words, I get a formal power series in $ct$, for some real number $c$ which cannot be determined. My problem is to determine the radius of convergence of this power series in $ct$.

Alas, my expertise with analytic solutions of ODEs stops with the standard undergraduate fare of the Frobenius method,... but only for linear equations. Hence I am asking the MO community for some readable reference(s) for the nonlinear case.

Thanks in advance.

Added (in response to comments below and particularly KConrad's answer)

What I am actually interested is in whether the solution will blow up in finite time. (Actually, in the problem $t$ is not really time, but inverse distance from a black-hole-like singularity and by blowing up in finite time, what I am after is whether the solution is indeed a black hole; i.e., whether there is an event horizon.) I cannot prove that this is the same as the formal power series solution having a finite radius of convergence, but this is precisely what happens in the black hole solutions I know: Schwarzschild and Reissner-Nordström, for instance.

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    $\begingroup$ Do you really want the radius of convergence, rather than the maximum existence time of the ODE? $\endgroup$ – Deane Yang May 28 '10 at 22:09
  • $\begingroup$ He really just wants maximum existence time. See his comment to my answer below. By the way, with the f_n's being polynomials, does the radius of convergence not necessarily equal the maximum existence time? $\endgroup$ – KConrad May 28 '10 at 22:14
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    $\begingroup$ Thanks, I missed what he said. I don't see why the radius of convergence should equal the maximum existence time. For example, all you have to do is cook up an ODE where the solution is, say, $x(t) = 1 - (1+t^2)^{-1}$. This has radius of convergence equal to 1 but infinite existence time. $\endgroup$ – Deane Yang May 28 '10 at 22:21
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    $\begingroup$ Deane: Oh, of course. The function x(t) = 1/(1+t^2) is defined for all t and satisfies x'(t) = -2t*x(t)^2 and x(0) = 1, with the power series having radius of convergence 1. Did you add a constant to the function for a specific reason? $\endgroup$ – KConrad May 28 '10 at 22:55
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    $\begingroup$ I found the texts "Ordinary differential equations in the complex domain" by Einar Hille, reissued by Dover, 1997 and "Ordinary differential equations" by E. L. Ince, reissued by Dover, 1956 (still available some time ago) rather helpful. $\endgroup$ – thomashennecke Apr 16 '14 at 8:50

Since the $f_n$'s are polynomials, does that tell us that the radius of convergence should equal a blow-up time (where the solution first diverges)? If the $f_n$'s were merely smooth I would be reluctant to suggest that. All I can pass along is a suggestion of how to get a bound on the blow-up time, which hopefully is equal to the radius of convergence: use differential inequalities.

An expert in non-linear ODEs told me once that in practice nobody tries to prove very sharp approximations for blow-up times; mere existence of a blow-up time usually suffices.

Here is an example. Consider $y'(t) = y(t)^2 - t$, with $y(0) = 1$. It has a solution which blows up in finite time. To estimate the blow-up time, let $Y(t) = 1/y(t)$ and see where $Y(t) = 0$. From $Y'(t) = tY(t)^2 - 1$ and $Y(0) = 1$, a computer algebra package has $Y(t) = 0$ at $t \approx 1.125$. We'd like to prove a theorem related to this numerical observation.

Claim: The solution to $y'(t) = y(t)^2 - t$ satisfying $y(0) = 1$ is undefined somewhere before $t = 1.221$.

Remark: This is weaker than what numerics suggest (i.e., the blow-up time is around 1.125), but proving something sharper requires a more careful analysis than I wish to develop.

Proof: We know $y(t)$ is defined for small $t > 0$. Assume $y(t)$ is defined for $0 \leq t < c$. We will show for an explicit number $c$ that $y(t) \geq c/(c-t)$ for $0 \leq t < c$, so $y(t) \rightarrow \infty$ as $t \rightarrow c^{+}$. Therefore $y(t)$ has to be undefined for some $t \leq c$.

Set $z(t) = c/(c-t)$, with $c$ still to be determined, so $$ \frac{\rm d}{{\rm d}t}(y - z) = y^2 - t - \frac{{\rm d}z}{{\rm d}t}. $$ After some algebra on the right, this becomes $$ \frac{\rm d}{{\rm d}t}(y - z) = (y - z)(y + z) + \frac{(c-1)c}{(c-t)^2} - t. $$

By calculus, $(c-1)c/(c-t)^2 - t \geq 0$ for $0 \leq t < c$ as long as $c - 1 \geq (4/27)c^2$, which happens for $c$ between the two roots of $x - 1 = (4/27)x^2$. The roots are approximately $1.2207$ and $5.5292$. So taking $c = 1.221$, we have $$(y(t) - z(t))' \geq (y(t)-z(t))(y(t)+z(t))$$ for $0 \leq t < c$. Using an integrating factor, this differential inequality is the same as $$ \frac{{\rm d}}{{\rm d}t}\left(e^{-\int_0^t(y(s)+z(s)){\rm d}s}(y(t) - z(t))\right) \geq 0. $$ Since $e^{-\int_0^t(y(s)+z(s)){\rm d}s}(y(t)-z(t))|_{t = 0} = 0$, $e^{-\int_0^t(y(s)+z(s)){\rm d}s}(y(t)-z(t)) \geq 0$ for $t \geq 0$, so $y(t) - z(t) \geq 0$ because the exponential factor is positive. Thus $y(t) \geq z(t) = c/(c-t)$. QED

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  • $\begingroup$ Thanks, your answer appears very helpful. Indeed, the question I'm interested in is precisely to determine whether there is a finite blow-up time or not. I just phrased in what I thought would be more "analytic" language. $\endgroup$ – José Figueroa-O'Farrill May 28 '10 at 21:31

If what you're after is really time of existence, I think your best bet is to study the equivalent integral equation

$$ x(t) = \Phi(x)(t), $$


$$ \Phi(x)(t) = x(0) + \int_0^t f(t, x(t))\, dt $$

and try to find a boundedness or growth condition on $x$ which implies the the same condition for $\Phi(x)$.

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  • $\begingroup$ Thanks. After I posted the question I asked a colleague and he suggested this very approach. Clearly I have a number of things to try now! $\endgroup$ – José Figueroa-O'Farrill May 28 '10 at 22:32

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