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Consider the following ODE: $$f'(r) = f^2(r) + O \left( \frac{1}{r^4} \right)$$ as $r$ goes to infinity. The initial conditions are $f(1) = C <0$.

What is the behaviour of a solution $f$ at infinity? (not only the leading term).

This ODE is equivalent to: $$w'' + O \left( \frac{1}{r^4} \right) w = 0$$

where $f = -\frac{w'}{w}$.

If I replace $O \left( \frac{1}{r^4} \right)$ with just $\frac{1}{r^4}$, then I know that the 2 linearly independent solutions are $f(r) = -\frac1r -\frac{1}{r^2} \tan{\frac1r}$ and $f(r) = -\frac1r + \frac{1}{r^2} \cot{\frac1r}$

Does that mean that the solution will always be $f(r) = -\frac1r + O \left( \frac{1}{r^3} \right)$ maybe for some family of initial conditions?

Any help is appreciated. If there is a reference that does things like that in detail, please share it with me.

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In your example you obtained two linearly independent solutions with different behavior: $\cot(1/r)\sim r,\; r\to\infty$, so your second solution is $O(r^{-2})$.

This is the general pattern if you assume that your perturbation is analytic at $\infty$. Write your equation as $$f'=f^2+q(r),\quad q(r)=r^{-4}\sum_{0}^\infty a_kr^{-k}.$$ When this is so, make the change of the variable $r=1/\zeta$, $y(\zeta)=w(1/\zeta)$ in your linear equation for $w$ ($f=-w'/w$) and obtain $$\zeta^2y''+2\zeta y'+\zeta^2(q_0+\ldots)y=0.$$ The indicial equation $\rho(\rho-1)+2\rho=0$ has roots $0,-1$, so solutions $y$ can be bounded or behave as $c/\zeta$ near $\zeta=0$. Returning to your original equation, this means that some solutions $f$ are like $-1/r$, while others are like $O(r^{-2})$.

Which is the case for a particular initial condition is impossible to decide because your restriction $O(r^{-4})$ tells nothing about $q$ on a long finite interval.

My argument shows that $O(r^{-4})$ can be relaxed to $O(r^{-3})$ with the same conclusion.

My assumption that $q$ is analytic at $\infty$ is of course too strong, and can be relaxed, depending on your needs, but I don't think it can be completely dropped.

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  • $\begingroup$ Thank you so much. Following this method, I am getting that one solution is $-\frac1r + O (r^{-3})$ and the other is as you said $O (r^{-2})$. Did you get that too? Also, to what can I relax analyticity at infinity so I can get the same conclusion? $\endgroup$
    – Laithy
    Commented Apr 22, 2019 at 21:52
  • $\begingroup$ never-mind, I think I made a mistake; the $O (r^{-3})$ should not be there. Also, Robert gave an example. $\endgroup$
    – Laithy
    Commented Apr 22, 2019 at 23:42
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Maybe for some family of initial conditions (depending on the $O(1/r^4)$ term). But note that $f(r) = -1/r + c/r^2$ is a solution to $f'(r) = f(r)^2 - c^2/r^4$ with initial condition $f(1) = c-1$.

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  • $\begingroup$ Thank you for your answer. Can we at least say that whatever the $ O \left(\frac{1}{r^4} \right )$ term is, we always have that $f$ would behave like $ -\frac{1}{r} + O \left(\frac{1}{r^{1+\epsilon}} \right)$ for some $\epsilon>0$ and maybe even find that $\epsilon$? How would we even prove that (if it's true)? $\endgroup$
    – Laithy
    Commented Apr 22, 2019 at 17:28

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