# Difficulty with Spectral Method using Chebyshev Polynomials

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?

• Perhaps you can link to the paper you are referencing? – Aron Ahmadia Aug 31 '12 at 8:59
• 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. – tau1777 Aug 31 '12 at 17:59
• Note that $y$ itself is a (linear) function of $\frac{r}{R}$ (and hence of $r$). – Christian Clason Nov 27 '12 at 17:08

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.