Derivatives Approximation on non uniform grid

Question

I was trying to approximate 1st derivative of a function $\phi$ on a non uniform grids: basically my aim is to do this on a uniform grid on the same domain, so I can calculate it on the "new" grid. For example:

I have $x_k$ a non uniform grid, and $\xi_k$ a uniform grid, both in $[0,1]$ and with the same length.

Now what I have is:

1- $\phi_k=\phi(x_k)$

2- $\{x_k\}, \{\xi_k\}$

and I want to know $\phi_x(x_k)$ with an High order approx.

Tried to do:

a) calculate an "high order approximation" using $\phi(x_k)$ and uniform grid formula (with central finite difference) --> call it $\phi^u_{x_k}$

b) do $\phi_x=\phi^u_{x_k}*\frac{d\xi}{dx}$

but no good.

Any idea?

Answer 1

So not quite sure what you're missing, but here's how you go about doing this sort of thing.

So first, I am assuming this is 1D due to your description. Second, I'm assuming you know the relationship between points in the physical domain, $x$, and the computational domain, $\xi$, something along the lines of $x=x(\xi)$. Given you have this relationship, you can do the following for some function $\phi(\cdot)$:

$$\begin{align} \frac{\partial \phi}{\partial x} &= \frac{\partial \phi}{\partial \xi} \frac{\partial \xi}{\partial x}\\ \frac{\partial \phi}{\partial x} &= \frac{\partial \phi}{\partial \xi} \left(\frac{\partial x}{\partial \xi}\right)^{-1}\\ \end{align}$$

With this representation, since you know $x(\xi)$, you can produce the second term exactly in the multiplication. So now you just need to approximate the first term, $\frac{\partial \phi}{\partial \xi}$. This term can be approximated using normal finite difference schemes. Using a central difference, for example, you could get the following:

$$\frac{\partial \phi}{\partial x} = \left(\frac{\phi_{i+1}-\phi_{i-1}}{2\Delta \xi}\right) \left(\frac{\partial x}{\partial \xi}\right)^{-1}$$

In this case, $\phi_{i}$ is associated to both $x_i$ and $\xi_i$, where $\phi_i = \phi(x_i)$, and where, in addition, $\phi_i = \phi(x(\xi_i))$. This should help you understand how to compute the necessary quantity you're after.