# Chebyshev spectral differentiation matrix for mapped domain

This is a follow up to this question.

I have mapped from Chebyshev space to the physical space using the mapping $x = ((1-w)\xi^3 - x_p\xi^2 + w\xi + x_p + 1 )*(L/2)$.

When in Chebyshev space I have easily defined the spectral differentiation matrix as follows:

$$(D_N)_{00} = \frac{2N^2 +1}{6}$$ $$(D_N)_{NN} = -\frac{2N^2 +1}{6}$$ $$(D_N)_{jj} = -\frac{\xi_j}{2(1-\xi_j^2)}, \ \forall \ 1 \leq j \leq N-1$$ $$(D_N)_{ij} = \frac{c_i}{c_j} \frac{(-1)^{i+j}}{\xi_i - \xi_j}, \ \forall \ i \neq j$$

where

$c_i= \begin{cases} 2,& \text{if } i = 0 \text{ or } N\\ 1, & \text{otherwise} \end{cases}$

$c_j= \begin{cases} 2,& \text{if } j = 0 \text{ or } N\\ 1, & \text{otherwise} \end{cases}$

1. How do I find the analogous differentiation matrix for the mapping
mentioned above?
2. Will this mapping affect the stability and convergence of collocation method?
• You can always choose to transform the ode / pde such the differentation matrices don't change. I can expand this to answer if you are interested. I think your transformation is already meant to be used that way.
– Bort
Jul 27, 2017 at 12:28
• @Bort : Please expand this as an answer Jul 27, 2017 at 12:37

Let's say you have derivatives of different order $u_x$ in your problem on your physical domain and a transformation $x=f(\xi)$.
The following example is taken from Boyd's book. Be the original ode: $$a_2(x)u_{xx}+a_1(x)u_{x}+a_0(x)u=g(x)$$ you solve now: $$a_2(f(\xi))\frac{f^\prime(\xi)u_{\xi\xi}-f^{\prime\prime}(\xi)u_\xi}{f^\prime(\xi)^3}+a_1(f(\xi))\frac{u_\xi}{f^\prime(\xi)}+a_0(f(\xi))u=g(f(\xi))$$ Now you can use the defined derivative operators for $\xi$.
This is essentially a variable transformation in your derivatives: $$\frac{du}{dx}=\frac{du}{d\xi}\frac{d\xi}{dx}=\frac{du}{d\xi}\frac{1}{\frac{dx}{d\xi}}=\frac{du}{d\xi}\frac{1}{f^\prime(\xi)}$$ This can of course be expanded to the multidimensional case, although it gets a bit more complicated.