I am having a bit of difficulty in trying to understand a paper. The paper uses spectral method to solve for an eigenvalue that comes from a system of coupled ODEs. I will write out only one equation now, because it is enough to get to the crux of my question(s).

The equation is

$V[r] = \frac{e^{-(\nu[r] +\lambda[r])}}{\epsilon[r] + p[r]} *\biggr[ (\epsilon[r] + p[r])( e^{\nu[r] +\lambda[r]})r W[r] \biggr]'$

I carry out the derivative and get

(Eq1) $V = \biggr[ \frac{\epsilon' +p'}{\epsilon + p} + r(\nu'+\lambda') +1 \biggr] W + r W'$

Now according to the paper I should be able to expand equilibrium quantities $(\epsilon ,p ,\nu ,\lambda$) of the system as Chebyshev Polynomials of the form

$B[r] = \Sigma_{i=0}^{\infty}b_i T_i[y] - \frac{1}{2} b_0 $, where $T_i[y]$ are the polynomials. I know how to get the $b_i$ using code I wrote in Mathematica. Also $y = 2(r/R) -1$, and the domain of $r$ is $(0,R)$.

The paper also states that the functions ($V,W$) can be expanded as $F[r] = (\frac{r}{R})^l \Sigma_{i=0}^{\infty}f_i T_i[y] - \frac{1}{2} f_0 $, and that in general a term like $B[r]F[r]$ can be expressed as

$B[r]F[r] = \frac{1}{2} (\frac{r}{R})^l \Sigma_{i=0}^{\infty} \pi_i T_i[y] - \frac{1}{2} \pi _0 $

where $\pi_i = \Sigma_{j=0}^{\infty}[b_{i+j} + \Theta(j-1)b_{|i-1|} ] f_l $ and $\Theta(k) = 0$ for $k<0$ and equals 1 for $k\geq 0$.

With all that being said, lets say I make the following equilibrium functions

$\frac{\epsilon' +p'}{\epsilon + p} = B_1[r] $ and $ r(\nu'+\lambda') = B_2[r]$, Then Eq1 becomes

(Eq2) $ \bigg((\frac{r}{R})^l \Sigma_{i=0}^{\infty}v_i T_i[y] - \frac{1}{2} v_0 \bigg) = \biggr[ B_1[r] + B_2[r] +1 \biggr] \biggr( (\frac{r}{R})^l \Sigma_{i=0}^{\infty}w_i T_i[y] - \frac{1}{2} w_0 \biggr) + r W' $.

Question1: What do I do with the $(\frac{r}{R})^l $terms? The polynomials are functions of $[y]$ so how can I even have an expansion like $B[r]= (\frac{r}{R})^l$ X function of [y]? Also it seems like I can just divide them out on each side of the equation, so what was the point of introduction that term? I mean, according to the paper this term is supposed to impose the boundary condition that $V,W$ go to zero as $r$ goes to zero.

*Question2:*How am I supposed to deal with the $r$ in the $r*W'$ term. The paper gives a description of how to handle derivative terms, but what about the $r$ itself. Am I supposed to treat it like an equilibrium value and use the rule for terms like $B[r]F[r]$ or should I express this $r$ in terms of $y$. Or should I do something else altogether?

  • 3
    $\begingroup$ Perhaps you can link to the paper you are referencing? $\endgroup$ Aug 31, 2012 at 8:59
  • $\begingroup$ Hi Aron, Here is the link arxiv.org/pdf/gr-qc/0210102.pdf The numerical things I'am having trouble with are desrcribed in Appendix A, and the equation I was examining above is equation (19). Thanks. $\endgroup$
    – tau1777
    Aug 31, 2012 at 17:59
  • $\begingroup$ Note that $y$ itself is a (linear) function of $\frac{r}{R}$ (and hence of $r$). $\endgroup$ Nov 27, 2012 at 17:08

1 Answer 1


I'm not sure it's possible to answer the question without a detailed reading of the paper. But in regards to the first question, you have $r/R = (y+1)/2$. And this factor cannot be divided out since it does not multiply all terms.

For question 2: since this equation is to be used to apply the boundary condition at $r=0$, I think the term you mention should vanish.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.