1
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 Apr 22 '19 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 Apr 22 '19 at 23:42
1
$\begingroup$

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 $f(1) = c-1$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 Apr 22 '19 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.