Chebyshev spectral differentiation matrix for mapped domain

Question

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?

Answer 1

Let's say you have derivatives of different order $u_x$ in your problem on your physical domain and a transformation $x=f(\xi)$.
Instead for solving on the original ode on the old domain, you solve the transformed ode on the new domain.

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.

Beware that you need to transform all boundary conditions as well.