# Asymptotic Behaviour of Solutions to a Riccati-type ODE with Small Forcing Term

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.

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.

• 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? – Laithy Apr 22 '19 at 21:52
• never-mind, I think I made a mistake; the $O (r^{-3})$ should not be there. Also, Robert gave an example. – Laithy Apr 22 '19 at 23:42

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 $$f(1) = c-1$$.

• 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)? – Laithy Apr 22 '19 at 17:28