# Asymptotic behaviour of solutions to system of ODEs

Let $$Y:(0,+\infty)\to\mathbb{R}^n$$ be a solution to the system of ODEs $$L[Y]=0,$$ where $$L$$ is a linear operator which behaves, in a neighbourhood of 0, as $$L[Y](r)\simeq-Y''(r)-\frac{1}{r}Y'(r)+\frac{1}{r^2}Y(r)+AY(r)$$ where $$A$$ is a constant $$n\times n$$ real matrix. I'm interested in understanding the behaviour in a neighbourhood of 0 of the kernel of $$L$$.

If $$A=0$$, then it's clear that any element of the kernel behaves, at main order near 0, like a linear combination of $$re_j, r^{-1}e_j\quad{j=1,\dots,n}$$ where $$e_j$$ is the $$j$$-th element of the standard basis of $$\mathbb{R}^n$$. I wonder if this is true as well when $$A\ne0$$.

In first place, we can surely approximate the behaviour of $$L$$ by setting all elements of the diagonal of $$A$$ to 0 (since it appears a term of size $$1/r^2$$ in the expression of $$L$$).

In general, if we assume that all the components of $$Y$$ have the same "size", then one may think that the conclusion still holds true, because formally $$\frac{1}{r^2}Y(r)+AY(r)\simeq \frac{1}{r^2}Y(r)$$ but I don't think that the assumption here is justified. Without this assumption the same conclusion may fail, suppose indeed that the first equation is given in the following form $$-y_1''-\frac{1}{r}y_1'+\frac{1}{r^2}y_1+y_2=0,$$ here I don't know if we have control on the size of $$y_2$$; if for instance $$y_2\simeq\frac{1}{r^3}y_1\qquad (\ast)$$ the behaviour near 0 of $$y_1$$ changes completely. On the other hand I don't know if it's possible to have a behaviour like $$(\ast)$$ in a system like this.

Is there any formal way to face this problem? Does it depend on the form of $$A$$? Any help is very appreciated.

• I am no expert here, but I believe it might be worth giving a look into [Coddington & Levinson, Theory of ordinary differential equations] chapters 4 and 5. I suspect that you're tryinf to deal with a singularity of the second kind. May 10 '20 at 20:52