# Behavior of differential equation as argument goes to zero

I'm trying to solve a coupled set of ODEs, but before attempting the full numerical solution, I would like to get an idea of what the solution looks like around the origin.

The equation at hand is:

$$y''_l - (f'+g')y'_l + \biggr[ \frac{2-l^2-l}{x^2}e^{2f} - \frac{2}{x}(f'+g') - \frac{2}{x^2} \biggr]y_l = \frac{4}{x}(f'+g')z_l$$

$y,f,g,z$ are all functions of $x$, which has the domain ($0, X_0$). If I specifically take the $l=2$ case I of course have

$$y''_2 - (f'+g')y'_2 + \biggr[ \frac{-4}{x^2}e^{2f} - \frac{2}{x}(f'+g') - \frac{2}{x^2} \biggr]y_2 = \frac{4}{x}(f'+g')z_2$$

To avoid issues with singularities, I multiply both sides of the equation by $x^2$, to get, lets call it EQ1.

$$x^2 y''_2 - x^2(f'+g')y'_2 + \biggr[ -4e^{2f} - x(f'+g') - 2 \biggr]y_2 = 4 x(f'+g')z_2$$

Now if by some magic I know by that as $x \rightarrow 0$, $y_l = x^{l+1}$ so that $y_2 = x^3$. How would I determine what the functions $z_2$ is around the origin?

Actually I already know the answer: $z_2$ should also go as $x^3$, but I have not been able to show it.

Any help is much appreciated.

My attempt at a solution: I have tried to keep things general so I expand $f[x] = 1 + f_1 x + f_2 x^2 +f_3 x^3$, and similarly $g[x] = 1 + g_1 x + g_2 x^2 +g_3 x^3$ where $f_1,f_2,f_3, g_1,g_2,g_3$ are constants. I have kept up to third order because I want to substitute $y_2 = x^3$, so I figured I should take the other functions to that order as well.

As for the $e^{2f}$ term, I use a truncated Taylor series for the exponential, $1+\frac{x^2}{2!} + \frac{x^3}{3!}$, into which I substitue my expansion of $f[x]$. After expanding everything out and eliminating terms of higher order than $x^3$ from the right hand side of EQ1, I get something like $constant*x^3$, while the left hand side is third order polynomial times $z_2$.

I really just don't know how to proceed. Should I have left higher order terms in the RHS, so that I could divide by a third order polynomial and still end up with $z_2 \propto x^3$. I don't know how this would work because on the RHS I had terms has high as $x^{12}$.

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In the formulas with $l=2$ you still have $y_l$ before the equality sign; I suppose this is a typo. // Do you really know that $y_2=x^3$ or is it $y_2=x^3 + \text{higher order terms}$? – user53153 Jan 11 '13 at 23:43
@Pavel, Thanks for pointing out the error. I have fixed the equations. I do know that y2 = x^3, there should not be any other terms. – tau1777 Jan 12 '13 at 0:33

Your equation EQ1 forces $z_2$ to have a quadratic term. Indeed, the left-hand side of EQ1 is $$4(1-e^2)x^3-4(f_1+2e^2f_1+g_1)x^4+O(x^5) \tag{1}$$ This is equated to $4x(f'+g')z_2$. Clearly, $z_2$ must include the term $\frac{1-e^2}{f_1+g_1}x^2$.
n:=7;
z:=(x^2*diff(y,x$2)-x^2*(diff(F,x)+diff(G,x))*diff(y,x)-(4*exp(2*F)+x*diff(F+G,x)+2)*y)/(4*x*diff(F+G,x)); series(z,x=0,4);  is the series expansion for$z_2$. By the way, you can number displayed equations with commands like \tag{1}, as I did above. - Thanks a bunch. I have been doing my computations in Mathematica and I will try reproduce your result. I really wish you had got$z_2 \propto x^3$since that this what they are saying in this paper: arxiv.org/pdf/gr-qc/0210102.pdf They mention this behavior in appendix A below equation A4, and the equation I wrote above is equation 20. I've made some obvious replacements in the variables. Thanks again. – tau1777 Jan 12 '13 at 6:39 My solution is as follows. I essentially tried to reproduce 5PM's work in Maple with some mathematica code. First I create the generic expansions of the functions$f$and$g$using f = Sum{ff[k]*x^k, {k,0,5}] g = Sum{gf[k]*x^k, {k,0,5}]  I had to create the variables "ff", and "gg" so as to avoid recursions limit errors in mathematica. Then I substituted these expansions into equation 1 above and then then divided the right hand side so get$z_2$by itself. Then I used Series[z2[r], {r,0,5}]  For which I got the following output. (I have cleaned it up a bit, having only kept terms of order three, and replacing the "ff" and "gg" with just "f" and "g".)$ z_2= -\frac {(e^{2 f_0} - 1 ) x^2} {f_1 + g_1} + \frac {1} {4} x^3 \biggr \{ \frac {2 (4e^{2 f_0} - 4)(f_2 + g_2))} {(f_1 + g_1)^2} - \frac {8 e^{2 f_0} f_1 + 5 f_1 + 5 g_1} {f_1 + g_1}\biggr \}$I esentially get the same result as 5PM. From this analysis it seems that$z_2$goes as$x^2$. The reason this disagrees with the paper I was reading, is that I forgot that physically speaking$f_0$is very close to zero, and so the x^2 term does go to zero, and so the physical answer is$ z_2= \frac {1} {4} x^3 \biggr \{\frac {8 e^{2 f_0} f_1 + 5 f_1 + 5 g_1} {f_1 + g_1}\biggr \}\$ .