# CFL condition in polar coordinates

In this question, I suggested that the Couran-Friedrichs-Lewy (CFL) condition for the wave equation in polar coordinates reads

$$C = 2c\frac{\Delta t}{\Delta r \Delta \phi} \leq C_\max \enspace ,$$

where $c$ is the phase speed. I suggested this from an intuitive point of view, and it worked in that example. Nevertheless, This is probably not right, And I could not find an expression for this case.

Question: What is the CFL condition in polar coordinates?

This question was asked before in Math.SE:

## 1 Answer

I was going to write a comment, but the equation seems to view better in answers..

I assume Von Neumann analysis is the proper approach to derive this equation, but a coordinate transformation from the cartesian CFL condition (I took from wikipedia) is not somehow equivalent? Specifically:

\begin{equation} \Delta t \sum_{i=1}^3 \frac{u_i}{\Delta x_i} = \Delta t \left( \frac{u_r}{\Delta r} + \frac{u_{\phi}}{r \Delta \phi} + \frac{u_z}{\Delta z} \right) < C_{max} \\ \end{equation}

• That is definitely a clever option, although that implies that a discretization that includes the origin is really problematic for the selection of $\Delta t$. – nicoguaro Mar 5 '17 at 20:58
• Regarding the von Neumann analysis, I thought about that, but how is the expansion? If one thinks to translate the analysis from the 1D case, it might involve a Fourier-Bessel expansion? Also, the concept of wave-number for polar coordinates is not as natural as it is in Cartesian ones. – nicoguaro Mar 5 '17 at 21:01
• I'm not sure how the expansion would pan out, or, for that matter if you'll be able to define a criterion without a transcendental function.. As for your first comment, though, I would argue that, analytically your argument is true but, the CFL condition should be measured at collocated cell centers, so the first radial element is located at $r=0+0.5 \Delta r$, meaning that there's no issue at the origin. Also, this analog is in agreement when $r$ is very large and the elements are nearly rectangular.. – Charles Mar 5 '17 at 22:01