# 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.

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)?

• This probably does belong on math.sx, and I would not think it's on topic here. It's not clear what equation you're starting from (e.g. you don't provide a definition for $l$), so that may be hindering your math.sx question. – E.P. Aug 25 '13 at 12:26
• I think, I'm not exactly sure, that l is just a parameter. I asked here because physicists know more on the topic of AdS3. – dingo_d Aug 25 '13 at 12:28
• That may be, but the content of your question is all mathematics. And again, without stating your original ODE this is very hard to answer - as is any question of the form "how do I derive this equation $...$?" that does not provide the necessary premises. – E.P. Aug 25 '13 at 12:37
• You could see i f i t gets answers on Math Overflow. – Abhimanyu Pallavi Sudhir Aug 25 '13 at 17:54
• @dingo_d Well, Math Overflow is generally only for research-level math. If your question is just needing an application of well-known ODE theory, it belongs on Math Stackexchange. If it calls upon physical principles in some meaningful way (beyond just "this came up while doing physics"), then it is okay here. But I doubt it will be particularly well-received on Math Overflow. In general, though, if you feel the question should be moved, you can ping a moderator and request they migrate it for you, to avoid cross-site duplication. – user43318 Aug 26 '13 at 1:01

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., $n\ge2$.