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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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Do you really want the radius of convergence, rather than the maximum existence time of the ODE? –  Deane Yang May 28 '10 at 22:09
    
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? –  KConrad May 28 '10 at 22:14
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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. –  Deane Yang May 28 '10 at 22:21
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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? –  KConrad May 28 '10 at 22:55
    
To make $x(0) = 0$, which was not needed. –  Deane Yang May 28 '10 at 22:57

2 Answers 2

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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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. –  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

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

where

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

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

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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! –  José Figueroa-O'Farrill May 28 '10 at 22:32

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