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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.

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    $\begingroup$ 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. $\endgroup$ May 10 '20 at 20:52

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