# Solving differential equations with sums (power series)

I have sets of 10 differential equations, but for this purpose I'll demonstrate what I need on one example that can be solved by hand.

My equation is this:

$$4GJ\Omega(\theta)\Omega'(\theta)\xi^\theta(\tau,r,\theta,\varphi)+4GJ\Omega^2(\theta)\frac{\partial \xi^\theta}{\partial\theta}=\mathcal{O}(r^{-1})$$

I need to assume that the solution of $\xi^\theta$ is in the form

$$\xi^\theta=\sum\limits_n\xi_n^\theta(\tau,\theta,\varphi)r^n$$

and it truncates for large n, where n is negative.

If I put that in my equation, and say that all powers greater or equal of 0 will be zero, I get

$$\Omega'\xi^\theta_n+\Omega \xi_{n,\theta}^\theta=0,\quad for\ n\geq0$$

where $\xi_{n,\theta}^\theta=\frac{\partial \xi_n^\theta}{\partial\theta}$ for shorthand notation simplification.

Now, this basically means that my general solution for $\xi^\theta$ is

$$\xi^\theta=\xi^\theta_{-1}r^{-1}+\mathcal{O}(r^{-2})$$

which is what I need to get.

Now, my question is: is it possible to do this for all other equations in Mathematica.

I got my equations, and I made substitution:

ξθ=Sum[ξ[τ, θ, φ][m]*r^m, m]


and got the result

4 G J Ω[θ]((Sum[D[ξ[τ, θ, φ][m],θ]*r^m, m])Ω[θ]+(Sum[ξ[τ, θ, φ][m]*r^m, m])Ω'[θ])


As I should. Is it possible now, to somehow get the most general result, like I got by solving by hand?

I'm asking because I have 10 equations like this, and some have other terms ($\xi^r,\xi^\tau,\xi^\varphi$) that I need to find the general solution to.

Do I actually expand it to some order and somehow truncate higher terms?

EDIT:

The power series solution is in the negative expansion.

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Related (single variable): math.stackexchange.com/questions/35591/… –  Michael E2 Oct 10 '13 at 12:13
Oops, that one is related but I meant this one: mathematica.stackexchange.com/questions/25363/… –  Michael E2 Oct 10 '13 at 12:20
I looked at the second one, but I don't have any boundary or initial conditions, other than that the series truncates :\ –  dingo_d Oct 10 '13 at 12:25
I tried with your 1st suggestion, but how do I expand this to negative exponents? –  dingo_d Oct 10 '13 at 12:33
Perhaps something like u = With[{lmax = 4}, SeriesData[r, Infinity, Array[a, lmax], 0, lmax, 1]]. You can think of series in 1/r as series at infinity. –  Michael E2 Oct 10 '13 at 12:48