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My question concerns coordinate maps and non-equally spaced fourier transforms.

I have dependent variables $(X(\xi),Y(\xi))$, where $\xi\in(0,2\pi)$. In general, $Y$ is assumed even and expanded as a Fourier series, ie

$$Y(\xi) = \sum_{k=-N}^{N} Y_k e^{-ik\xi},$$

where $N$ is taken to tend towards infinity and the reality condition implies $Y_{-k} = Y_k$

Now, my governing equations are derived from a lagrangian. I won't go into the full details of it, since they are anscillary for the question, but it is illustrative to look at the potential energy term, V, which takes the form

$$V=\int_0^{2\pi} Y(\xi)^2 \frac{dX(\xi)}{d\xi} \ d\xi,$$

with $X-\xi$ being the Hilbert transform of $Y$ (because (X,Y) are real and imaginary parts of an analytic function in $\mathbb{C}$). That is, $$X = \xi + \sum i \sigma_k Y_k e^{-ik\xi},$$

with $\sigma_i$ being 1 for $i>0$, 0 when $i=0$, and -1 when $i<0$.

Substituting in the Fourier expansion, $V$ can be written entirely in terms of the $Y_k$s. The Euler-Larange equations then returns a set of governing (algebraic) equations are found and a system of equations is solved for a given $N$. It turns out that for the solutions I desire, the series converges very slowly, and this has to do with the fact that $Y(\xi)$ is very localized.

After going through Boyd's book (in particular, see chapter 16), it seems like one way to deal with this is to map the independent variable into a different, non-uniformly spaced map. One example is to send $$\xi\to \theta-sin(\theta)$$ (Boyd calls this a Kepler mapping), for $\theta \in(0,2\pi)$, which will tend to crowd more points towards the ends of the domain, which is where my functions are localized.

This seems promising to me, and I would like to pursue it.

To this end, I take a change of coordinates at the level of the integral $V$, shown above, to find

$$V = \int_0^{2\pi} Y(\theta)^2 \frac{d X(\theta)}{d \theta} \ d\theta$$.

From here it is unclear to me how one takes a Fourier type expansion (needed to allow for the easy computation of the Hilbert transform) on this non-uniform grid.

Any suggestions would be greatly appreciated,

Nick

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Unless I'm missing something obvious, if you transform $Y=Y_\xi(\xi)$ to use $Y=Y_\theta(\theta) = Y_\xi(\theta-\sin\theta)$, then the function $w(\xi)=X(\xi)+iY(\xi)$, which, as you say, was analytic in $\xi$, would be analytic in $\theta$ also, with $w(\theta) = w(\xi=\theta-\sin\theta)$, so that whatever Fourier series $Y(\theta)$ has w.r.t. $\theta$, $X$ would have the same expression as before in terms of the new variable $\theta$ and the new Forier coefficients $Y_k^{(\theta)}$ (it's a generic expression, independent of the coordinates being used to write it down).

This is assuming that the rest of your equations can be rewritten in terms of $\theta$ as well: I don't know a good way to relate two Fourier series obtained by a change of variable. But this is not the same as doing a discrete Fourier transform on a non-uniform grid, anyway, because you seem to be using a Galerkin-type method, and evaluating all the necessary integrals in closed form by hand. In fact, I don't quite see where in your question a discrete Fourier transform is being used.

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  • $\begingroup$ Thanks for the reply. I think I follow your reasoning, and in fact this is what my computations were showing. However, doesn't this imply that the coefficients, ie the $Y_k$, are invariant to the choice of coordinates, as the governing equations remain invariant? This does not make sense to me. $\endgroup$ – Nick P Feb 13 '15 at 3:56
  • $\begingroup$ @NickP No, of course not. The $Y_k$ are different, you have one set $Y_k^{(\xi)}$ coming from $Y(\xi)$, and another set $Y_k^{(\theta)}$. Same function, different variable, therefore different sets of Fourier coefficients. One way to think of them is to note that $e^{ik\xi} = e^{ik\theta}e^{-ik\sin\theta}$, so to relate the two sets of coefficients you can try expanding $e^{-ik\sin\theta}$ in Fourier series. $\endgroup$ – Kirill Feb 13 '15 at 4:16
  • $\begingroup$ Ah, yes. I see now. Thanks for the advice (and patience)! $\endgroup$ – Nick P Feb 13 '15 at 8:58

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