Solving ODE with negative expansion power series Since on Mathematics stackexchange I didn't get an answer, I'll try it here, since people here are more familiar with this topic (general relativity related).
I am reading a dissertation of Porfyriadis "Boundary Conditions, Effective Action, and Virasoro Algebra for $AdS_3$", and I am trying to solve a system of DE to get the appropriate diffeomorfism (page 31 onward).
I am trying to solve a system of ODE, such that each DE is equal to some degree of term that I'm expanding to. For instance, one DE is this:
$\xi^r\partial_r g_{rr}+2g_{tt}\partial_t\xi^t=\mathcal{O}(r)$
which you get by taking a Lie derivative of background metric and setting it equal to certain $\mathcal{O}(r^n)$ terms.
$g_{ij}$ are given, from metric ofc. I need to assume that the solution (since I'm looking for components of $\xi^\mu$( which is a vector with components $\xi^t, \xi^r, \xi^\phi$) is given with power series of the form:
$\xi^\mu=\sum\limits_{n}\xi^\mu_n(t,\phi)r^n$,
and this is to be seen as expansion around 1/r (expansion around $r=\infty$).
Now when I plug this in the ODE I get this
$\frac{2}{l^2}\sum_n\xi^r_nr^{n+1}+2\sum_n\xi^t_{n,t} r^n+\frac{2}{l^2}\sum_n\xi^t_{n,t}r^{n+2}=\mathcal{O}(r)$, where
$\xi^\mu_{n,i}$
is the derivative with the respect to i-th component.
What troubles me is, how to expand this? Do I set n=0,-1,-2,... until my O(r) terms cancel each other out? Or?
I'm kinda stuck, at this seemingly easy point.
In the thesis he gets 6 equations with coefficients, first one should be:
$\xi^r_{n-1}+l^2\xi^t_{n,t}+\xi^t_{n-2,t}=0,\ n\ge 2$,
But I am not getting this. What am I doing wrong?

EDIT: For further clarity: The metric is that of $AdS_3$ given with line element:
$ds^2=-\left(1+\frac{r^2}{l^2}\right)dt^2+\left(1+\frac{r^2}{l^2}\right)^{-1}dr^2+r^2 d\phi^2$,
and the differential equations in question are given by solving $\mathcal{L}_\xi g_{\mu\nu}=\mathcal{O}(r^n)$, where $\mathcal{O}(r^n)$ are the fall off conditions. In the dissertation, he took the deviation of nonzero components of the metric to be subleading, that is:
$\mathcal{L}_\xi g_{tt}=\mathcal{O}(r)$
$\mathcal{L}_\xi g_{rr}=\mathcal{O}(r^{-3})$
$\mathcal{L}_\xi g_{\phi\phi}=\mathcal{O}(r)$, while others are $\mathcal{O}(1)$.
Solving Lie derivative gives me 6 equations, which I should solve by plugging in the above ansatz ($\xi^\mu=\sum\limits_{n}\xi^\mu_n(t,\phi)r^n$), but this is the part I get stuck.
ADDENDUM:
I was looking at other components, and have noticed that I have $g_{rr}$ factor with some of them. That term in metric is:
$g_{rr}=\left(1+\frac{r^2}{l^2}\right)^{-1}$
Now, is it legitimate thing to expand this around $r=\infty$ so that I can put $r$ terms inside the sums (assumed solution)?
 A: I didn't really read the question (at least the stuff about metrics or whatever), but I will write what I think is the answer anyway. You are doing an expansion around $r=\infty$. You have $$\frac{2}{l^2}\sum_n\xi^r_nr^{n+1}+2\sum_n\xi^t_{n,t} r^n+\frac{2}{l^2}\sum_n\xi^t_{n,t}r^{n+2}=\mathcal{O}(r).$$ This equation says that as $r \to \infty$, the LHS, which is $\frac{2}{l^2}\sum_n\xi^r_nr^{n+1}+2\sum_n\xi^t_{n,t} r^n+\frac{2}{l^2}\sum_n\xi^t_{n,t}r^{n+2}$ only goes to $\infty$ as fast as $r$ to the first power. This means that the higher powers of $r$ from the three terms cancel each other. 
Well how can they cancel each other? Lets look at the $r^2$ piece. The first term gives a contribution $\frac{2}{l^2}\xi^r_1r^{2}$. The second term gives a contribution $2\xi^t_{2,t} r^2$, and the third term gives a contribution $\frac{2}{l^2}\xi^t_{0,t}r^{2}$. The sum of these contributions is $$\frac{2}{l^2}\xi^r_1r^{2}+2\xi^t_{2,t} r^2+\frac{2}{l^2}\xi^t_{0,t}r^{2} = (\frac{2}{l^2}\xi^r_1r^{2}+2\xi^t_{2,t}+\frac{2}{l^2}\xi^t_{0,t})r^2.$$ For this to be zero, we must have $$\frac{2}{l^2}\xi^r_1+2\xi^t_{2,t}+\frac{2}{l^2}\xi^t_{0,t} =0,$$ or rearranging, $$\xi^r_1+l^2\xi^t_{2,t}+\xi^t_{0,t} =0.$$
If that is good let's move on to the general $n\ge 2$. The first term gives a contribution $\frac{2}{l^2}\xi^r_{n-1}r^{n}$. The second term gives a contribution $2\xi^t_{n,t} r^n$, and the third term gives a contribution $\frac{2}{l^2}\xi^t_{n-2,t}r^{n}$. The sum of these contributions is $$\frac{2}{l^2}\xi^r_{n-1}r^{n}+2\xi^t_{n,t} r^n+\frac{2}{l^2}\xi^t_{n-2,t}r^{n} = (\frac{2}{l^2}\xi^r_{n-1}r^{2}+2\xi^t_{n,t}+\frac{2}{l^2}\xi^t_{n-2,t})r^n.$$ For this to be zero, we must have $$\frac{2}{l^2}\xi^r_{n-1}+2\xi^t_{n,t}+\frac{2}{l^2}\xi^t_{n-2,t} =0,$$ or rearranging, $$\xi^r_{n-1}+l^2\xi^t_{n,t}+\xi^t_{n-2,t} =0.$$ The $n$ where this needed to be zero were the $n$ for higher than linear terms, i.e., $n\ge2$. 
This is the equation that you said you didn't understand how they got it. I hope this clears up part of the confusion.
