# Solving ODE with negative expansion power series

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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 post imported from StackExchange Mathematics at 2014-06-09 18:51 (UCT), posted by SE-user dingo_d
retagged Jun 9, 2014
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.

This post imported from StackExchange Mathematics at 2014-06-09 18:51 (UCT), posted by SE-user episanty
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.

This post imported from StackExchange Mathematics at 2014-06-09 18:51 (UCT), posted by SE-user dingo_d
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.

This post imported from StackExchange Mathematics at 2014-06-09 18:51 (UCT), posted by SE-user episanty
You could see i f i t gets answers on Math Overflow.

This post imported from StackExchange Mathematics at 2014-06-09 18:51 (UCT), posted by SE-user dimensio1n0
Didn't even know of Math Overflow :\ So I should post there? I'll give it a try...

This post imported from StackExchange Mathematics at 2014-06-09 18:51 (UCT), posted by SE-user dingo_d
Well, yes, in the end, the question is just about ODEs, so maybe it won't be appropriate on MO. '

This post imported from StackExchange Mathematics at 2014-06-09 18:51 (UCT), posted by SE-user dimensio1n0
There is a meta question regarding the closing of this question here. Basically, the argument is that this should not be a duplicate as it has an answer, while the "original" question does not. So the duplicate chain does not make sense!

This post imported from StackExchange Mathematics at 2014-06-09 18:51 (UCT), posted by SE-user user1729 PhD

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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.

This post imported from StackExchange Mathematics at 2014-06-09 18:51 (UCT), posted by SE-user NowIGetToLearnWhatAHeadIs
answered Aug 28, 2013 by (80 points)
Finally someone who know how to explain it!!!!! THANK YOU SO MUCH!! I'll try to set up a bounty and give it to you, you deserve it :)

This post imported from StackExchange Mathematics at 2014-06-09 18:51 (UCT), posted by SE-user dingo_d
Oh and, while I'm at it, do you have any material on this kind of solving method? If it has any special name so that I can google it or sth like it? I'd be really grateful, since I looked all over but had no luck...

This post imported from StackExchange Mathematics at 2014-06-09 18:51 (UCT), posted by SE-user dingo_d
I don't know of a name for it besides just using power series. There probably is a name. The only other time I remember seeing something quite like this is solving the quantum harmonic oscillator using power series. Here is a link:Quantum Harmonic Oscillator. Maybe someone else will be more helpful.

This post imported from StackExchange Mathematics at 2014-06-09 18:51 (UCT), posted by SE-user NowIGetToLearnWhatAHeadIs
I'll look it up, maybe I find a proper name for this :D I'll give you bounty in an hour (at least that's what M.SE says :D)

This post imported from StackExchange Mathematics at 2014-06-09 18:51 (UCT), posted by SE-user dingo_d

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